Writing a linear function?? Precalculus I am suppose write a linear function for $h(x)$ and I am given $h(6) = -3$ and $h(2) = 7$
Iam not asking u to do the problem for me just a link or a little help thanks
 A: So, you are giving two points : $(6, -3)$ and $(2,7)$ living on your line. To find the equation of such a line, you need the main ingredient: the slope $m$
$$ m = \frac{7 - (-3)}{2 - 6} = \frac{10}{-4} = \frac{-5}{2}$$
$$ \therefore h(x) - 7 = \frac{-5}{2}(x -2) \implies h(x) = \frac{-5x}{2} + 5+ 7$$
$$ \therefore h(x) = - \frac{5x}{2} + 12 $$
is the required line
A: Since the slope between any two points is constant you may write down an equation of the line immediately:
$$\frac{y+3}{x-6}=\frac{7+3}{2-6}.$$  
A: There is another approach for this problem. Since $h(x)$ is a linear function, it assumes the form of $h(x)=mx+b$ which is in the slope-intercept form. It can be shown that the slope is $m=-5/2$. We are given that $h(2)=7$ and $h(6)=-3$, so we can use either of this given information to find the value of $b$. I will use in this case the given information $h(2)=7$. $h(2)=2(-5/2)+b$ which implies that $7=2(-5/2)+b=-5+b$. Solving algebraically for $b$, yields $b=12$. Therefore, the linear function that satisfies the initial conditions is $h(x)=-5x/2 + 12$
