Question

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

Answer 1

`"vertex "=(-3/2,-49/4),x=-5,x=2`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form "` is.

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form use "color(blue)"completing the square"`

`y=x^2+2(3/2)xcolor(red)(+9/4)color(red)(-9/4)-10`

`color(white)(y)=(x+3/2)^2-49/4`

`color(magenta)"vertex "=(-3/2,-49/4)`

`"to find the x-intercepts let y = 0"`

`x^2+3x-10=0`

`"the factors of - 10 which sum to + 3 are + 5 and - 2"`

`(x+5)(x-2)=0`

`"equate each factor to zero and solve for "x`

`x=-5,x=2larrcolor(red)"x-intercepts"`