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

1 Answer
Jim G.
Jun 22, 2018

#"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"#

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