Mathematics » Functions III » Quadratic Functions

Inverse of the Function y = ax2

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

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

To determine the inverse function of $$y=ax^{2}$$:

$\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay^{2} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & \cfrac{x}{a} = y^{2} \\ &&\therefore y = ±\sqrt{\cfrac{x}{a}} \qquad (x \geq 0) \end{array}$

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.

• 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.

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.

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*}

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*}

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*}

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

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

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 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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