How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=-2x^2+8x-3#?

1 Answer
Shwetank Mauria
May 28, 2017

#y#-intercept is #f(0)=-3#, axis of symmetry is #x-2=0# and vertex is #(2,5)#

Explanation:

The vertex form of a quadratic equation is #y=f(x)=a(x-h)^2+k#, where vertex is #(h,k)# and axis of symmetry is #x-h=0# and #y#-intercept is given by #f(0)#.

Writing #f(x)=-2x^2+8x-3# in vertex form, we get

#f(x)=-2x^2+8x-3#

#=-2(x^2-4x+4)+8-3#

#=-2(x-2)^2+5#

Hence #y#-intercept is #f(0)=-3#

axis of symmetry is #x-2=0#

and vertex is #(2,5)#

graph{(y+2x^2-8x+3)(x-2)((x-2)^2+(y-5)^2-0.02)(x^2+(y+3)^2-0.02)=0 [-8.705, 11.295, -3.72, 6.28]}

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