How do you find the VERTEX of a parabola #f(x)= -x^2 + 3x + 10#?

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Answer 1 · 2015-07-17T22:12:17

Find vertex of# f(x) = -x^2 + 3x + 10#
Vertex #(3/2, 49/4)#

x coordinate of vertex: #x = (-b/2a) = -3/-2 = 3/2#

y coordinate of vertex: #y = f(3/2) = - 9/4 + 9/2 + 10 = 49/4#

Answer 2 · 2015-07-17T22:31:47

#x=3/2 or 1 1/2#

#y=49/4 or 12 1/4#

The vertex is #(3/2, 49/4)# or #(1 1/2, 12 1/4)#.

#f(x)=-x^2+3x+10#

Substitute #y# for #f(x)#.

#y=-x^2+3x+10# is in the form #ax^2+bx+c#, where #a=-1, b=3, c=10#.

To find the value of #x#, use the equation #x=(-b)/(2a)#.

#x=(-3)/(2*-1)#

#x=(-3)/(-2)# =

#x=3/2=1 1/2#

To find the #y# value, substitute #3/2# for #x# in the equation #y=x^2+3x+10#.

#y=-(3/2)^2+3(3/2)+10# =

#y=-9/4+9/2+10#

The common denominator is #4#. Multiply the terms on the right side by the fraction that will give each term a denominator of #4#.

#y=-9/4+9/2*2/2+10*4/4#

#y=-9/4+18/4+40/4# =

#y=49/4=12 1/4#.