How do you find the vertex and intercepts for $y = 10x – 3x^2$ ?

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Answer 1 · 2016-06-17T16:28:52

The vertex is $(5/3,25/3)$
The intercepts are (0,0) and $(10/3,0)$

you can find the coordinate x of the vertex using the formula
$x=-b/(2a)$
where a is the coefficient of $x^2$ and b of x

Since a=-3 and b=10, you have

$x=-10/-6=5/3$

Then you can put this value in x and find y
$y=10(5/3)-3(5/3)^2$

$y=50/3-3(25/9)$
$y=50/3-25/3$
$y=25/3$

One of the intercepts is the origin (0,0) because y has no known term

By solving the equation $10x-3x^2=0$ you can find the other one:

$x(10-3x)=0$

$10-3x=0$

$x=10/3$

How do you find the vertex and intercepts for y = 10x – 3x^2y = 10x – 3x^2?