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 $k$?
Asked
Active
Viewed 70 times
1
-
1Have you tried to write $x=3,$ $y=10$ and see what happens? – mfl Jul 07 '14 at 20:24
-
6Plug in, and it's over. – André Nicolas Jul 07 '14 at 20:24
2 Answers
4
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: $$\rm\underset{\underset{\displaystyle y}{\displaystyle\uparrow}}{10}=k(\underset{\underset{\displaystyle x}{\displaystyle\uparrow}}3-1)^2\ \ \Rightarrow \ \ 10=4k \ \ \Rightarrow \ \ k=\dfrac52.\,\checkmark$$
Hakim
- 10,213
1
Since the point $(x,y)=(3,10)$ satisfies the equation we can substitute into the equation:
$$10=k(3-1)^2$$
$$10=4k\Rightarrow k=\frac{5}{2}$$