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Suppose that the line tangent to the graph of $y = h(x)$ at $x = 3$ passes through the points $(-2, 3)$ and $(4, -1)$ with a slope of $-2/3$. Find $h(3)$.

Hey guys, here's a question from my test review and I know how to do it, I know the answer, but I'm confused as to the "why" not the "how". Specifically, why should my mind jump straight to "Use slope-intercept form to find the value at $h(3$)"?

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What your mind should jump to is that the tangent line to a graph of a function $h(x)$ at $x=a$ must pass through the graph at that $x$ value, which means it must pass through the point $(a,h(a))$. Then because you know the line should pass through the point $(3,h(3))$, your job is to find the $y$ such that $(3,y)$ is on the graph of the line.

You are given more than enough information to determine the $y$ value on the line when the $x$ value is $3$, and "slope-intercept form" is just one way to organize your search. Eg., another way would be to use the notion of slope directly and work forward or backwards from your points, going down $2/3$ when $x$ goes up $1$, or up $2/3$ when $x$ goes down $1$.

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