Solve a linear function How do I solve this homework assignment?
For a linear function $y=f(x)$, $f(-3) = 25$ and $f(3) = 11$.
Determine $f(-20)$.
I know that with the values $f(-3) = 25$ and $f(3) = 11$ I am suppose to derive a formula and the plug in $f(-20)$ but how do I derive a formula from the values given?
Thanks!!
 A: Hint: You don't have to derive the formula. Use the fact that the gradient of a linear function is always constant.
$$\frac{f(-20) - f(3)}{-20 - 3} = \frac{f(3) - f(-3)}{3 - (-3)}$$
Of course, if there is a need, you can derive similarly the formula of $f$. This is, by the way, my favourite method of deriving the equation of a linear function defined by two given points.
$$\frac{f(x) - f(3)}{x - 3} = \frac{f(3) - f(-3)}{3 - (-3)}$$
A: Strictly speaking, the function $f(x)=ax+b$ is actually affine, rather linear. Affinity in this case implies that 
$$
f(\alpha x+(1-\alpha)y)=\alpha f(x)+(1-\alpha)f(y).
$$
With $\alpha=\frac{23}{6}$, you have $\alpha(-3)+(1-\alpha)3=-20$, so
$$f(-20)=f\left(\frac{23}{6}(-3)+\frac{-17}{6}(3)\right)=\frac{23}{6}f(-3)+\frac{-17}{6}f(3)=\frac{194}{3}.
$$
A: The slope is $m=\frac{25-11}{-3-3}=\frac{14}{-6}=-2\frac13$.
Now, the $y$-intercept is: $b=y-mx=25-(-2\frac13)(-3)=25-7=18$.
Hence, the formula for the line is: $f(x)=-2\frac13x+18$.
Plug in $x=-20$ to get $f(-20)=-2\frac13(-20)+18=46\frac23+18=\boxed{64\frac23}$.
A: Going from $-3$ to $3$ ($6$ units more), the function value changes by $11-25=-14$ units. So the function changes by $-14/6=-7/3$ for an increment of $1$ unit.
Going from $-3$ to $-20$, i.e. a move of $-17$ units, the function value changes by $(-17)\cdot(-7/3)$ and reaches $25+119/3=194/3$.
CHECK:
Going from $3$ to $-20$, i.e. a move of $-23$ units, the function value changes by $(-23)\cdot(-7/3)$ and reaches $11+161/3=194/3$.
GENERALIZATION:
Going from $3$ to $x$, i.e. a move of $x-3$ units, the function value changes by $(x-3)\cdot(-7/3)$ and reaches $11-7/3(x-3)$.
A: For a linear function, if you know its value at two points, you can fully determine the function. Do you know that general formula for a straight line in the plane? Write that down, and plug in (-3,25) and (3,11) to solve for the coefficients.
A: the fact that $f$ is linear means, in this context, that
$$
f(x) = (1-x)f(0) + xf(1) \tag{1}
$$
(since this is obviously satisfied for $x=0,1$ and a line is determined by any two distinct points)
replacing $x$ with $-x$ gives
$$
f(-x) = (1+x)f(0) - xf(1)
$$
so by addition
$$
f(x)+f(-x) = 2f(0)
$$
for $x= 3$ this gives 
$$
11+25=2f(0)
$$
i.e. $f(0) = 18$
by substitution of either $\pm3$ in (1) we find that $f(1)=\frac{47}3$. knowing the values of $f(0)$ and $f(1)$ allows you to find the value for any $x$ by substitution in (1)
