Mathematics » Functions III » Quadratic Functions

Inverse of the Function y = ax2

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

Example

Question

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} \\ \cfrac{x}{3} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{3}} \qquad (x \geq 0) \end{align*}

Sketch the graphs on the same system of axes

Inverse of the Function <em>y = ax<sup>2</sup></em>

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

Inverse of the Function <em>y = ax<sup>2</sup></em>

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

\[\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay^{2} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & \cfrac{x}{a} = y^{2} \\ &&\therefore y = ±\sqrt{\cfrac{x}{a}} \qquad (x \geq 0) \end{array}\]Inverse of the Function <em>y = ax<sup>2</sup></em>

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 \(y = ±\sqrt{\cfrac{x}{3}}\):

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

    Inverse of the Function <em>y = ax<sup>2</sup></em>

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

    Inverse of the Function <em>y = ax<sup>2</sup></em>

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.

Inverse of the Function <em>y = ax<sup>2</sup></em>

Example

Question

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} \\ \cfrac{x}{7} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{7}} \qquad (x \geq 0) \end{align*}

Sketch both graphs on the same system of axes

Inverse of the Function <em>y = ax<sup>2</sup></em>

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 \(( q^{-1} )\). 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*}Inverse of the Function <em>y = ax<sup>2</sup></em>

or

Option \(\text{2}\): Restrict the domain of \(q\) to \(x \le 0\) so that the inverse will also be a function \(( q^{-1} )\). 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*}

Inverse of the Function <em>y = ax<sup>2</sup></em>

Example

Question

  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

Inverse of the Function <em>y = ax<sup>2</sup></em>

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*}

Determine \(f\).

\begin{align*} \text{Let } y & = – \cfrac{1}{2} \sqrt{x} \\ \text{Interchange } x \text{ and } y: \quad x & = – \cfrac{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*} – \cfrac{1}{4} x & = 4x^{2} \\ 0 & = 4x^{2} + \cfrac{1}{4} x \\ 0 & = x (4x + \cfrac{1}{4} ) \\ \therefore x = 0 & \text{ or } 4x + \cfrac{1}{4} = 0 \\ \text{If } x = 0: \enspace y &= 0 \\ \text{If } 4x + \cfrac{1}{4} &= 0: \\ 4x &= – \cfrac{1}{4} \\ x &= – \cfrac{1}{16} \\ \text{If } x = – \cfrac{1}{16}: \enspace y &= – \cfrac{1}{4} ( – \cfrac{1}{16} ) \\ &= \cfrac{1}{64} \end{align*}

Therefore, the two graphs intersect at \(( 0;0 )\) and \(( – \cfrac{1}{16}; \cfrac{1}{64} )\).

Example

Question

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) \\ \cfrac{x}{2} & = y^{2} \\ y & = \sqrt{\cfrac{x}{2}} \qquad (x \ge 0, y \geq 0) \\ & \\ \therefore h^{-1}(x) & = \sqrt{\cfrac{x}{2}} \qquad (x \geq 0) \end{align*}

Determine the point of intersection

\begin{align*} 2x^{2} & = \sqrt{\cfrac{x}{2}} \\ ( 2x^{2} )^{2} & = ( \sqrt{\cfrac{x}{2}} )^{2} \\ 4x^{4} & = \cfrac{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} & = \cfrac{1}{8} \\ \therefore x & = \cfrac{1}{2} \\ \text{If } x= \cfrac{1}{2}, \quad y & = \cfrac{1}{2} \end{align*}

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

Sketch both graphs on the same system of axes

Inverse of the Function <em>y = ax<sup>2</sup></em>

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 \(B( \cfrac{1}{2} ; \cfrac{1}{2})\).

\begin{align*} \text{Average gradient: } &= \cfrac{y_{B} – y_{A}}{x_{B} – x_{A}} \\ &= \cfrac{ \cfrac{1}{2} – 0 }{\cfrac{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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