SAT math question regarding slope

Question

In the xy-plane, the graph of $y=k(x-1)^2$, where $k$ is a constant, passes through the point $(3,10)$. What is the value of $k$?

Answer 1

That the graph of $\,\rm y=k(x-1)^2\,$ passes through the point with coordinates $(3,10)$ means that the pair $(3,10)$ satisfies that equation. So if you plug in those values you'll get: $$\rm\underset{\underset{\displaystyle y}{\displaystyle\uparrow}}{10}=k(\underset{\underset{\displaystyle x}{\displaystyle\uparrow}}3-1)^2\ \ \Rightarrow \ \ 10=4k \ \ \Rightarrow \ \ k=\dfrac52.\,\checkmark$$

Answer 2

Since the point $(x,y)=(3,10)$ satisfies the equation we can substitute into the equation:

$$10=k(3-1)^2$$

$$10=4k\Rightarrow k=\frac{5}{2}$$