How do you find the vertex and intercepts for y+12x-2x^2=15?

1 Answer
Dec 12, 2017

"see explanation"

Explanation:

"express in standard form"

rArry=2x^2-12x+15

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

color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))

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

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

• " the coefficient of the "x^2" term must be 1"

rArr2(x^2-6x)+15

• " add/subtract "(1/2"coefficient of x-term")^2"to"
x^2-6x

2(x^2+2(-3)xcolor(red)(+9)color(red)(-9))+15

=2(x-3)^2-18+15

rArry=2(x-3)^2-3larrcolor(red)"in vertex form"

rArrcolor(magenta)"vertex "=(3,-3)

color(blue)"Intercepts"

• " let x = 0, in equation for y-intercept"

• " let y = 0, in equation for x-intercepts"

x=0toy=2(-3)^2-3=15larrcolor(red)"y-intercept"

y=0to2(x-3)^2-3=0

rArr(x-3)^2=3/2

color(blue)"take the square root of both sides"

rArrx-3=+-sqrt(3/2)larrcolor(blue)"note plus or minus"

rArrx=3+-sqrt(3/2)larrcolor(red)"x-intercepts"

rArrx~~1.78,x~~4.22" to 2 dec. places"