# Suppose that f is a linear function such that f(3) = 6 and f(-2) = 1. What is f(8)?

Jan 23, 2018

$f \left(8\right) = 11$

#### Explanation:

Since it's a linear function, it must be of the form

$a x + b = 0 \text{ " " } \left(1\right)$

So

$f \left(3\right) = 3 a + b = 6$

$f \left(- 2\right) = - 2 a + b = 1$

Solving for $a$ and $b$ gives $1$ and $3$, respectively.

Therefore, substituting the values of $a$, $b$, and $x = 8$ in equation $\left(1\right)$ gives

$f \left(8\right) = 1 \cdot 8 + 3 = 11$

Jan 23, 2018

$f \left(8\right) = 11$

A lot more explanation is involved than doing the actual maths

#### Explanation:

Linear basically means 'in line'. This is implying a strait line graph situation You read left to right on the x-axis so the first value is the least $x$
using:

$f \left(- 2\right) = {y}_{1} = 1$
$f \left(3\right) = {y}_{2} = 6$
$f \left(8\right) = {y}_{3} = \text{Unknown}$

Set point 1 as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 2 , 1\right)$
Set point 2 as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(3 , 6\right)$
Set point 2 as ${P}_{3} \to \left({x}_{3} , {y}_{3}\right) = \left(8 , {y}_{3}\right)$

The gradient (slope) of part will be the same gradient of the whole.

Gradient (slope) is the amount of up or down for a given amount of along, reading left to right.

Thus the gradient gives us: ${P}_{1} \to {P}_{2}$

$\left(\text{change in "y)/("change in } x\right) \to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{6 - 1}{3 - \left(- 2\right)} = \frac{5}{5}$

Thus we have ${P}_{1} \to {P}_{3}$ ( same ratio )

$\left(\text{change in "y)/("change in } x\right) \to \frac{{y}_{3} - {y}_{1}}{{x}_{3} - {x}_{1}} = \frac{{y}_{3} - 1}{8 - \left(- 2\right)} = \frac{5}{5}$

$\textcolor{w h i t e}{\text{dddddddd")-> color(white)("ddd")(y_3-y_1)/(x_3-x_1) =color(white)("d")(y_3-1)/10color(white)("d}} = 1$

Multiply both sides by 10

$\textcolor{w h i t e}{\text{dddddddd")->color(white)("dddddddddddddd")y_3-1color(white)("d}} = 10$

$\textcolor{w h i t e}{\text{dddddddd")->color(white)("ddddddddddddddddd")y_3color(white)("d}} = 11$