How do you find the vertex of #f(x)=-2x^2+5x-7#?

1 Answer
Nov 14, 2017

The vertex is #(5/4,-31/8)#.

Explanation:

#f(x)=-2x^2+5x-7# is a quadratic equation in standard form:

#f(x)=ax^2+bx+c#,

where:

#a=-2#, #b=5#, and #c=-7#

The axis of symmetry is the #x#-value for the vertex. The formula for the axis of symmetry is #x=(-b)/(2a)#.

#x=(-5)/(2*-2)#

#x=(-5)/(-4)#

#x=5/4#

To find the #y#-value of the vertex, substitute #y# for #f(x)#, and #5/4# for #x#.

#y=-2(5/4)^2+5(5/4)-7#

#y=-2(25/16)+25/4-7#

#y=-50/16+25/4-7#

Simplify #-50/16# to #-25/8#.

#y=-25/8+25/4-7#

Multiply each fraction by a fraction equal to #1#, so that the denominator is #8#. For example, #6/6=1#.

#y=-25/8+(25/4xxcolor(red)(2/2))-(7xxcolor(blue)(8/8))#

#y=-25/8+50/8-56/8#

#y=-31/8#

The vertex is #(5/4,-31/8)#, which is also #(1.25,-3.875)#. Because #a<0#, the vertex is the maximum point and the parabola opens downward.

graph{y=-2x^2+5x-7 [-14, 14.47, -11.39, 2.85]}