How do you write a quadratic function in intercept form whose graph has x intercepts -5, 5 and passes through (6, 11)?

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Answer 1 · 2017-04-06T06:35:42

Quadratic function is #y=f(x)=x^2-25#

As the quadratic function, say #y=f(x)# has #x#-intercepts #-5, 5#

it's value is #0#, when #x=5# and #x=-5#.

Hence equation must be of the type

#y=f(x)=a(x-5)(x-(-5))=a(x-5)(x+5)#

as it passes trough #(6,11)#, we have

#11=a(6-5)(6+5)=axx1xx11=11a#

Hence, #a=1# and quadratic function is #y=f(x)=(x-5)(x+5)=x^2-25#