How do you find the VERTEX of a parabola y=x^2-3x-10?

1 Answer
May 14, 2018

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

Explanation:

"given a parabola in "color(blue)"standard form "color(white)(x)ax^2+bx+c

"then the x-coordinate of the vertex is"

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

y=x^2-3x-10" is in standard form"

"with "a=1,b=-3" and "c=-10

rArrx_("vertex")=-(-3)/2=3/2

"substitute this value into the equation for y"

y_("vertex")=(3/2)^2-3(3/2)-10=-49/4

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