How do you find the vertex and the intercepts for #f(x)= -6x^2+ 5x + 18#?

1 Answer
Jim G.
Jul 31, 2018

#"vertex "=(5/12,457/24),x=5/12+-sqrt457/12#

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 "color(blue)"complete the square"#

#y=-6(x^2-5/6x-3)#

#color(white)(y)=-6(x^2+2(-5/12)x+25/144-25/144-3)#

#color(white)(y)=-6(x-5/12)+457/24larrcolor(blue)"in vertex form"#

#color(magenta)"vertex "=(5/12,457/24)#

#"to obtain the x-intercepts set y = 0"#

#-6(x-5/12)^2+457/24=0#

#(x-5/12)^2=457/144#

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

#x-5/12=+-sqrt(457/144)=+-sqrt457/12#

#"add "5/12" to both sides"#

#x=5/12+-sqrt457/12larrcolor(red)"exact values"#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1405 views around the world
You can reuse this answer
Creative Commons License