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Quadratic Functions

2.5 Quadratic functions (EMCFC)

Inverse of the function \(y=a{x}^{2}\) (EMCFD)

Worked example 6: Inverse of the function \(y=a{x}^{2}\)

Determine the inverse of the quadratic function \(h(x) = 3x^{2}\) and sketch both graphs on the same system of axes.

Determine the inverse of the given function \(h(x)\)

  • Interchange \(x\) and \(y\) in the equation.
  • Make \(y\) the subject of the new equation.
\begin{align*} \text{Let } y & = 3x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 3y^{2} \\ \frac{x}{3} & = y^{2} \\ \therefore y & = ±\sqrt{\frac{x}{3}} \qquad (x \geq 0) \end{align*}

Sketch the graphs on the same system of axes

dd68fe550a0df4adff14611b7d466fcd.png

Notice that the inverse does not pass the vertical line test and therefore is not a function.

6b7c091c3560a1613be248811eb623c6.png

To determine the inverse function of \(y=ax^{2}\):

\[\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay^{2} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & \frac{x}{a} = y^{2} \\ &&\therefore y = ±\sqrt{\frac{x}{a}} \qquad (x \geq 0) \end{array}\] 0070380311ea74d751ab4a3e986977ff.png

The vertical line test shows that the inverse of a parabola is not a function. However, we can limit the domain of the parabola so that the inverse of the parabola is a function.

Domain and range

Consider the previous worked example \(h(x) = 3x^{2}\) and its inverse \(y = ±\sqrt{\frac{x}{3}}\):

  • If we restrict the domain of \(h\) so that \(x\ge 0\), then \(h^{-1}(x) = \sqrt{\frac{x}{3}}\) passes the vertical line test and is a function.

    c7d92df29bbfa65300159461eeba607c.png
  • If the restriction on the domain of \(h\) is \(x\le 0\), then \(h^{-1}(x) = -\sqrt{\frac{x}{3}}\) would also be a function.

    a18dace9eabe2d8de12c765f7db89a86.png

The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.

Similarly, a restriction on the domain of the function results in a restriction on the range of the inverse and vice versa.

ca08fee60fd68f1fa993cdff88536c33.png

Worked example 7: Inverses - domain, range and restrictions

Determine the inverse of \(q(x) = 7x^{2}\) and sketch both graphs on the same system of axes. Restrict the domain of \(q\) so that the inverse is a function.

Examine the function and determine the inverse

Determine the inverse of the function:

\begin{align*} \text{Let } y & = 7x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 7y^{2} \\ \frac{x}{7} & = y^{2} \\ \therefore y & = ±\sqrt{\frac{x}{7}} \qquad (x \geq 0) \end{align*}

Sketch both graphs on the same system of axes

ac074adc29c5cb82cfa51e20614e3a03.png

Determine the restriction on the domain

Option \(\text{1}\): Restrict the domain of \(q\) to \(x \ge 0\) so that the inverse will also be a function \(\left( q^{-1} \right)\). The restriction \(x \ge 0\) on the domain of \(q\) will restrict the range of \(q^{-1}\) such that \(y \ge 0\).

\begin{align*} q: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \end{align*}d96b13d3493aae67a5b0ae4810218421.png

or

Option \(\text{2}\): Restrict the domain of \(q\) to \(x \le 0\) so that the inverse will also be a function \(\left( q^{-1} \right)\). The restriction \(x \le 0\) on the domain of \(q\) will restrict the range of \(q^{-1}\) such that \(y \le 0\).

\begin{align*} q: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \end{align*}4523b04b83cbeff72ab25026259f9fb2.png

Worked example 8: Inverses - domain, range and restrictions

  1. Determine the inverse of \(f(x) = -x^{2}\).
  2. Sketch both graphs on the same system of axes.
  3. Restrict the domain of \(f\) so that its inverse is a function.

Determine the inverse of the function

\begin{align*} \text{Let } y & = -x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = -y^{2} \\ -x & = y^{2} \\ y & = ±\sqrt{-x} \qquad (x \leq 0) \end{align*}

Note: \(\sqrt{-x}\) is only defined if \(x \le 0\).

Sketch both graphs on the same system of axes

137eddb8a4cffa9e464f73cfb8ccf63f.png

The inverse does not pass the vertical line test and is not a function.

Determine the restriction on the domain

  • If \(f(x) = -x^{2}, \text{ for } x \le 0\): \begin{align*} f: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \end{align*}
  • If \(f(x) = -x^{2}, \text{ for } x \ge 0\): \begin{align*} f: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \end{align*}

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Inverses - domain, range, intercepts, restrictions

Exercise 2.4

Determine the inverse for each of the following functions:

\(y = \frac{3}{4}x^{2}\)
\begin{align*} y & = \frac{3}{4}x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = \frac{3}{4}y^{2} \\ \frac{4}{3}x & = y^{2} \\ \therefore y & = \pm \sqrt{\frac{4}{3}x} \qquad (x \geq 0) \end{align*}
\(4y - 8x^{2} = 0\)
\begin{align*} \text{Interchange } x \text{ and } y: \quad 4x - 8y^{2} & = 0 \\ 4x & = 8y^{2} \\ \frac{1}{2}x & = y^{2} \\ \therefore y & = \pm \sqrt{\frac{1}{2}x} \qquad (x \geq 0) \end{align*}
\(x^{2} + 5y = 0\)
\begin{align*} \text{Interchange } x \text{ and } y: \quad y^{2} + 5x & = 0 \\ y^{2} & = - 5x \\ \therefore y & = \pm \sqrt{- 5x} \qquad (x \leq 0) \end{align*}
\(4y - 9 = (x + 3)(x - 3)\)
\begin{align*} 4y - 9 &= x^{2} - 9 \\ 4y &= x^{2} \\ \text{Interchange } x \text{ and } y: \quad 4x & = y^{2} \\ \therefore y & = \pm \sqrt{4x} \qquad (x \geq 0) \end{align*}

Given the function \(g(x) = \frac{1}{2}x^{2}\) for \(x \ge 0\).

Find the inverse of \(g\).

\begin{align*} \text{Let } y & = \frac{1}{2}x^{2} \qquad (x \geq 0) \\ \text{Interchange } x \text{ and } y: \quad x & = \frac{1}{2}y^{2} \qquad (y \geq 0) \\ 2x & = y^{2} \\ y & = \sqrt{2x} \qquad (x \geq 0,y \geq 0) \\ & \\ \therefore g^{-1}(x) & = \sqrt{2x} \qquad (x \geq 0) \end{align*}

Draw \(g\) and \(g^{-1}\) on the same set of axes.

4846dd41dc26e0ba9c33bf1b7280157b.png

Is \(g^{-1}\) a function? Explain your answer.

Yes. It passes the vertical line test and is a one-to-one relation.

State the domain and range for \(g\) and \(g^{-1}\).

\begin{align*} g: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \\ g^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \end{align*}

Determine the coordinates of the point(s) of intersection of the function and its inverse.

To find the points of intersection, we equate \(g(x)\) and \(g^{-1}(x)\):

\begin{align*} \frac{1}{2}x^{2} & = \sqrt{2x} \\ \left( \frac{1}{2}x^{2} \right)^{2} & = \left( \sqrt{2x} \right)^{2} \\ \frac{1}{4}x^{4} & = 2x \\ x^{4} & = 8x \\ x^{4} - 8x & = 0\\ x(x^{3} - 8) & = 0\\ x(x-2)(x^{2} + 2x + 4) & = 0\\ \therefore x = 0 \text{ or } x =2 \text{ or } & x^{2} + 2x + 4 = 0 \end{align*} \begin{align*} \text{If } x & = 0 \\ y & = 0 \\ \text{If } x & = 2 \\ y & = \frac{1}{2}(2)^{2} \\ & = 2 \end{align*} \begin{align*} \text{If } x^{2} + 2x + 4 & = 0 \\ \text{Use the quadratic formula } x & = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\ & = \frac{-2 \pm \sqrt{(2)^{2} - 4(1)(4)}}{2(1)} \\ & = \frac{-2 \pm \sqrt{-12}}{2} \\ & = \text{no real solution} \end{align*}

Therefore, we get the points \((0;0)\) or \((2;2)\).

Given the graph of the parabola \(f(x) = ax^{2}\) with \(x \ge 0\) and passing through the point \(P(1;-3)\).

aff9d67a6f3229726a5e607571b886b4.png

Determine the equation of the parabola.

\begin{align*} \text{Let } y & = ax^{2} \\ \text{Substitute } P(1;-3): \quad & -3 = a(1)^{2} \\ \therefore a & = -3 \\ \therefore f(x) & = -3x^{2} \quad (x \ge 0) \end{align*}

State the domain and range of \(f\).

\(f: \enspace \text{domain } \{x: x \ge 0 \} \quad \text{range } \{y: y \le 0 \}\)

Give the coordinates of the point on \(f^{-1}\) that is symmetrical to the point \(P\) about the line \(y=x\).

\((-3;1)\)

Determine the equation of \(f^{-1}\).

\begin{align*} \text{Let } y & = -3x^{2} \qquad (x \geq 0) \\ \text{Interchange } x \text{ and } y: \quad x & = -3y^{2} \qquad (y \geq 0) \\ -\frac{1}{3}x & = y^{2} \\ y & = \sqrt{-\frac{1}{3}x} \qquad (x \leq 0,y \geq 0) \\ & \\ \therefore f^{-1}(x) & = \sqrt{-\frac{1}{3}x} \qquad (x \leq 0) \end{align*}

State the domain and range of \(f^{-1}\).

\(f^{-1}: \enspace \text{domain } \{x: x \le 0 \} \quad \text{range } \{y: y \ge 0 \}\)

Draw a graph of \(f^{-1}\).

ef1d4eaf0dd76b03309a61aa1e110a93.png

Determine the inverse of \(h(x) = \frac{11}{5}x^{2}\).

\begin{align*} \text{Let } y & = \frac{11}{5}x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = \frac{11}{5}y^{2} \\ \frac{5}{11} & = y^{2} \\ \therefore y & = ±\sqrt{\frac{5}{11}x} \qquad (x \geq 0) \end{align*}

Sketch both graphs on the same system of axes.

2b29ce512a2fc15515e02bc82a8dc0e8.png

Restrict the domain of \(h\) so that the inverse is a function.

Option \(\text{1}\): Restrict the domain of \(h\) to \(x \ge 0\) so that the inverse will also be a function \(\left( h^{-1} \right)\). The restriction \(x \ge 0\) on the domain of \(h\) will restrict the range of \(h^{-1}\) such that \(y \ge 0\).

\begin{align*} h: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \\ h^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \end{align*} ce5df8254379071177010bf41109c752.png

or

Option \(\text{2}\): Restrict the domain of \(h\) to \(x \le 0\) so that the inverse will also be a function \(\left( h^{-1} \right)\). The restriction \(x \le 0\) on the domain of \(h\) will restrict the range of \(h^{-1}\) such that \(y \le 0\).

\begin{align*} h: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \\ h^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \end{align*} 64c76e7cd612c20f2a98d683c7221943.png

The diagram shows the graph of \(g(x) = mx + c\) and \(f^{-1}(x) = a \sqrt{x}, \enspace (x \geq 0)\). Both graphs pass through the point \(P(4;-1)\).

c996080a36463bb349ad286e126c8b5b.png

Determine the values of \(a\), \(c\) and \(m\).

From the graph we see that the straight line passes through the origin, therefore \(c=0\).

\begin{align*} g(x) & = mx \\ \text{Substitute } P(4;-1): \enspace -1 & = 4m \\ \therefore m & = - \frac{1}{4} \\ \therefore g(x) & = - \frac{1}{4} x \\ & \\ f^{-1}(x)& = a \sqrt{x} \\ \text{Substitute } P(4;-1): \enspace -1 & = a \sqrt{4} \\ -1 & = 2a \\ \therefore a & = - \frac{1}{2} \\ \therefore f^{-1}(x) & = - \frac{1}{2} \sqrt{x} \end{align*}

Give the domain and range of \(f^{-1}\) and \(g\).

\begin{align*} g: \qquad & \text{domain: } x \in \mathbb{R} \quad \text{range: } y \in \mathbb{R} \\ f^{-1}: \qquad & \text{domain: } x \ge 0 \quad \text{range: } y \le 0 \end{align*}

For which values of \(x\) is \(g(x) < f(x)\)?

\(x > 4\)

Determine \(f\).

\begin{align*} \text{Let } y & = - \frac{1}{2} \sqrt{x} \\ \text{Interchange } x \text{ and } y: \quad x & = - \frac{1}{2} \sqrt{y} \qquad (y \geq 0) \\ -2x & = \sqrt{y} \\ \therefore y & = 4x^{2} \qquad (y \geq 0) \end{align*}

Determine the coordinates of the point(s) of intersection of \(g\) and \(f\) intersect.

To determine the coordinates of the point(s) of intersection, we equate \(g\) and \(f\):

\begin{align*} - \frac{1}{4} x & = 4x^{2} \\ 0 & = 4x^{2} + \frac{1}{4} x \\ 0 & = x \left(4x + \frac{1}{4} \right) \\ \therefore x = 0 & \text{ or } 4x + \frac{1}{4} = 0 \\ \text{If } x = 0: \enspace y &= 0 \\ \text{If } 4x + \frac{1}{4} &= 0: \\ 4x &= - \frac{1}{4} \\ x &= - \frac{1}{16} \\ \text{If } x = - \frac{1}{16}: \enspace y &= - \frac{1}{4} \left( - \frac{1}{16} \right) \\ &= \frac{1}{64} \end{align*}

Therefore, the two graphs intersect at \(\left( 0;0 \right)\) and \(\left( - \frac{1}{16}; \frac{1}{64} \right)\).

Worked example 9: Inverses - average gradient

Given: \(h(x) = 2x^{2}, \quad x \ge 0\)

  1. Determine the inverse, \(h^{-1}\).
  2. Find the point where \(h\) and \(h^{-1}\) intersect.
  3. Sketch \(h\) and \(h^{-1}\) on the same set of axes.
  4. Use the sketch to determine if \(h\) and \(h^{-1}\) are increasing or decreasing functions.
  5. Calculate the average gradient of \(h\) between the two points of intersection.

Determine the inverse of the function

\begin{align*} \text{Let } y & = 2x^{2} \qquad (x \ge 0) \\ \text{Interchange } x \text{ and } y: \quad x & = 2y^{2} \qquad (y \ge 0) \\ \frac{x}{2} & = y^{2} \\ y & = \sqrt{\frac{x}{2}} \qquad (x \ge 0, y \geq 0) \\ & \\ \therefore h^{-1}(x) & = \sqrt{\frac{x}{2}} \qquad (x \geq 0) \end{align*}

Determine the point of intersection

\begin{align*} 2x^{2} & = \sqrt{\frac{x}{2}} \\ \left( 2x^{2} \right)^{2} & = \left( \sqrt{\frac{x}{2}} \right)^{2} \\ 4x^{4} & = \frac{x}{2} \\ 8x^{4} & = x \\ 8x^{4} - x & = 0 \\ x(8x^{3} - 1) & = 0 \\ \therefore x = 0 &\text{ or } \enspace 8x^{3} - 1 = 0 \\ \text{If } x=0, \quad y & = 0 \\ \text{If } 8x^{3} - 1 & = 0 \\ 8x^{3} & = 1 \\ x^{3} & = \frac{1}{8} \\ \therefore x & = \frac{1}{2} \\ \text{If } x= \frac{1}{2}, \quad y & = \frac{1}{2} \end{align*}

Therefore, this gives the points A\((0;0)\) and \(B\left( \frac{1}{2} ; \frac{1}{2}\right)\).

Sketch both graphs on the same system of axes

2499f938e2c2c73b845e404cade5e859.png

Examine the graphs

From the graphs, we see that both \(h\) and \(h^{-1}\) pass the vertical line test and therefore are functions.

\begin{align*} h: \quad & \text{as } x \text{ increases, } y \text{ also increases, } \text{ therefore } h \text{ is an increasing function. } \\ h^{-1}: \quad & \text{as } x \text{ increases, } y \text{ also increases, } \text{ therefore } h^{-1} \text{ is an increasing function. } \end{align*}

Calculate the average gradient

Calculate the average gradient of \(h\) between the points A\((0;0)\) and \(B\left( \frac{1}{2} ; \frac{1}{2}\right)\).

\begin{align*} \text{Average gradient: } &= \frac{y_{B} - y_{A}}{x_{B} - x_{A}} \\ &= \frac{ \frac{1}{2} - 0 }{\frac{1}{2} - 0 } \\ &= 1 \end{align*}

Note: this is also the average gradient of \(h^{-1}\) between the points \(A\) and \(B\).

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Inverses - average gradient, increasing and decreasing functions

Exercise 2.5

Sketch the graph of \(y={x}^{2}\) and label a point other than the origin on the graph.

31f5a975b303c43e7d339f9175b9749b.png

Find the equation of the inverse of \(y = x^{2}\).

\begin{align*} y & = x^{2} \\ \text{Inverse: } x & = y^{2} \\ \therefore y & = \pm \sqrt{x} \qquad (x \ge 0) \end{align*}

Sketch the graph of the inverse on the same system of axes.

0135bb3b293155599b9b9a2fbb855518.png

Is the inverse a function? Explain your answer.

No. For certain values of \(x\), the inverse cuts a vertical line in two places. Therefore, it is not function.

\(P(2;4)\) is a point on \(y = x^{2}\). Determine the coordinates of \(Q\), the point on the graph of the inverse which is symmetrical to \(P\) about the line \(y = x\).

\(Q(4;2)\)

Determine the average gradient between:

  1. the origin and \(P\);
  2. the origin and \(Q\).

Interpret the answers.

  1. \begin{align*} \text{Average gradient: } &= \frac{y_{P} - y_{O}}{x_{P} - x_{O}} \\ &= \frac{ 4 - 0 }{2- 0 } \\ &= 2 \end{align*}
  2. \begin{align*} \text{Average gradient: } &= \frac{y_{Q} - y_{O}}{x_{Q} - x_{O}} \\ &= \frac{ 2 - 0 }{4- 0 } \\ &= \frac{1}{2} \end{align*}

\(\text{Average gradient}_{ OP} = 2\) and \(\text{average gradient}_{ OQ} = \frac{1}{2}\).

Both gradients are positive, and they are also reciprocals of each other.

Given the function \(f^{-1}(x) = kx^{2}\), \(x \ge 0\), which passes through the point \(P \left( \frac{1}{2}; -1 \right)\).

d659342d42004a0cceffe17e61c18c6f.png
Find the value of \(k\).
\begin{align*} f^{-1}(x) & = kx^{2} \\ \text{Substitute } \left( \frac{1}{2}; -1 \right) \quad -1 & = k \left( \frac{1}{2} \right)^{2} \\ -1 & = k \left( \frac{1}{4} \right) \\ -4 & = k \\ \therefore f^{-1}(x) & = -4x^{2} \end{align*}

State the domain and range of \(f^{-1}\).

\begin{align*} \text{Domain: } & \{x: x \ge 0, x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \le 0, y \in \mathbb{R} \} \end{align*}

Find the equation of \(f\).

\begin{align*} f^{-1}: \quad y & = -4x^{2} \quad (x \ge 0) \\ f: \quad x & = -4y^{2} \quad (y \ge 0) \\ -\frac{1}{4}x & = y^{2} \\ \therefore y & = \sqrt{-\frac{1}{4}x} \quad (x \le 0) \end{align*}

State the domain and range of \(f\).

\begin{align*} \text{Domain: } & \{x: x \le 0, x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \ge 0, y \in \mathbb{R} \} \end{align*}

Sketch the graphs of \(f\) and \(f^{-1}\) on the same system of axes.

158a370fe01383b82888f5778819181a.png

Is \(f\) an increasing or decreasing function?

Decreasing function: as the value of \(x\) increases, the function value decreases.

Given: \(g(x) = \frac{5}{2}x^{2}, \enspace x \ge 0\).

Find \(g^{-1}(x)\).

\begin{align*} \text{Let } y & = \frac{5}{2}x^{2} \qquad (x \ge 0) \\ \text{Interchange } x \text{ and } y: \quad x & = \frac{5}{2}y^{2} \qquad (y \ge 0) \\ \frac{2}{5}x & = y^{2} \\ y & = \sqrt{\frac{2}{5}x} \qquad (x \ge 0, y \geq 0) \\ & \\ \therefore g^{-1}(x) & = \sqrt{\frac{2}{5}x} \qquad (x \geq 0) \end{align*}

Calculate the point(s) where \(g\) and \(g^{-1}\) intersect.

\begin{align*} \frac{5}{2}x^{2} & = \sqrt{\frac{2}{5}x} \\ \left( \frac{5}{2}x^{2} \right)^{2} & = \left( \sqrt{\frac{2}{5}x} \right)^{2} \\ \frac{25}{4}x^{4} & = \frac{2}{5}x \\ \frac{25}{4}x^{4} - \frac{2}{5}x & = 0 \\ 125x^{4} - 8x & = 0 \\ x(125x^{3} - 8) & = 0 \\ x(5x - 2)(25x^{2} +10x + 4) & = 0 \\ \therefore x = 0 \enspace \text{ or } \enspace 5x - 2 = 0 \enspace &\text{ or } \enspace 25x^{2} +10x + 4 = 0 \\ \text{If } x=0, \quad y & = 0 \\ \text{If } x = \frac{2}{5}, \quad y & = \frac{5}{2} \left( \frac{2}{5} \right) ^{2} \\ \therefore y & = \frac{2}{5} \\ \text{If } 25x^{2} +10x + 4 &=0 \\ \text{Use quadratic formula: } \quad x &= \frac{-10 \pm \sqrt{100 - 4(25)(4)}}{2(25)}\\ &= \frac{-10 \pm \sqrt{-300}}{50} \\ \therefore & \text{no real solution} \end{align*}

Therefore, the points of intersection are \((0;0)\) and \(\left( \frac{2}{5};\frac{2}{5} \right)\).

Sketch \(g\) and \(g^{-1}\) on the same set of axes.

c0d4553686702d87367538bafc5b7bc4.png

Use the sketch to determine if \(g\) and \(g^{-1}\) are increasing or decreasing functions.

\begin{align*} g: \enspace & \text{as } x \text{ increases, } y \text{ also increases, } \therefore g \text{ is an increasing function. } \\ g^{-1}: \enspace & \text{as } x \text{ increases, } y \text{ also increases, } \therefore g^{-1} \text{ is an increasing function. } \end{align*}

Calculate the average gradient of \(g^{-1}\) between the two points of intersection.

\begin{align*} \text{Average gradient: } &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{ \frac{2}{5} - 0 }{\frac{2}{5} - 0 } \\ &= 1 \end{align*}