# 6.1 Parabolic and other functions

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## Investigation : average gradient - parabolic function

Fill in the table by calculating the average gradient over the indicated intervals for the function $f\left(x\right)=2x-2$ :

 ${x}_{1}$ ${x}_{2}$ ${y}_{1}$ ${y}_{2}$ $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ A-B B-C C-D D-E E-F F-G

What do you notice about the average gradient over each interval? What can you say about the average gradients between A and D compared to the averagegradients between D and G?

The average gradient of a parabolic function depends on the interval and is the gradient of a straight line that passes through the points on the interval.

For example, in [link] the various points have been joined by straight-lines. The average gradients between the joined points are then the gradients of the straight lines that pass through the points.

Given the equation of a curve and two points ( ${x}_{1}$ , ${x}_{2}$ ):

1. Write the equation of the curve in the form $y=...$ .
2. Calculate ${y}_{1}$ by substituting ${x}_{1}$ into the equation for the curve.
3. Calculate ${y}_{2}$ by substituting ${x}_{2}$ into the equation for the curve.
4. Calculate the average gradient using:
$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

Find the average gradient of the curve $y=5{x}^{2}-4$ between the points $x=-3$ and $x=3$

1. Label the points as follows:

${x}_{1}=-3$
${x}_{2}=3$

to make it easier to calculate the gradient.

2. We use the equation for the curve to calculate the $y$ -value at ${x}_{1}$ and ${x}_{2}$ .

$\begin{array}{ccc}\hfill {y}_{1}& =& 5{x}_{1}^{2}-4\hfill \\ & =& 5{\left(-3\right)}^{2}-4\hfill \\ & =& 5\left(9\right)-4\hfill \\ & =& 41\hfill \end{array}$
$\begin{array}{ccc}\hfill {y}_{2}& =& 5{x}_{2}^{2}-4\hfill \\ & =& 5{\left(3\right)}^{2}-4\hfill \\ & =& 5\left(9\right)-4\hfill \\ & =& 41\hfill \end{array}$
3. $\begin{array}{ccc}\hfill \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}& =& \frac{41-41}{3-\left(-3\right)}\hfill \\ & =& \frac{0}{3+3}\hfill \\ & =& \frac{0}{6}\hfill \\ & =& 0\hfill \end{array}$
4. The average gradient between $x=-3$ and $x=3$ on the curve $y=5{x}^{2}-4$ is 0.

## Average gradient for other functions

We can extend the concept of average gradient to any function. The average gradient for any function also depends on the interval chosen and is the gradient of a straight line that passes through the two points. So we can use the formula that we found for the average gradient of parabolic functions and apply it to any function. We will consider the average gradient of just two functions here: exponential functions and hyperbolic functions.

## Average gradient of exponential functions

For example, if we were asked to find the average gradient of the function $\mathrm{g\left(x\right)}=3.{2}^{x}+2$ between the points $\left(-4;\mathrm{2,2}\right)$ and $\left(-0,6;4\right)$ . This is shown in [link] .

Using the formula we find:
$\begin{array}{ccc}\hfill \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}& =& \frac{4-2,2}{\left(-0,6\right)-\left(-4\right)}\hfill \\ & =& \frac{1,8}{-0,6+4}\hfill \\ & =& \frac{1,8}{5,2}\hfill \\ & =& 0,35\hfill \end{array}$

## Average gradient of hyperbolic functions

For example, if we were asked to find the average gradient of the function $g\left(x\right)=\frac{2}{x}+2$ between the points $\left(-4;-2,5\right)$ and $\left(0,5;6\right)$ and $\left(-4;2,2\right)$ and $\left(-0,6;4\right)$ . This is shown in [link] .

For the first point we would get:

$\begin{array}{ccc}\hfill \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}& =& \frac{\left(-2,5\right)-1}{\left(-4\right)-0,5}\hfill \\ & =& \frac{-3,5}{-4,5}\hfill \\ & =& 0,78\hfill \end{array}$
Similarly for the second points we would find that the average gradient is: $0,53$

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I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
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20/(×-6^2)
Salomon
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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Abhi
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