How do you write a quadratic function in intercept form who has x intercepts -3, 7 and passes through points (6,-9)?

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Answer 1 · 2017-04-04T07:30:53

#y=(x+3)(x-7)#

Let us have a rough diagram
Look at the graph
Let the equation be

#y=(x-a)(x-b)#

Given #x = -3, 7#

The eqauation becomes -

#y=(x-(-3))(x-7)#
#y=(x+3)(x-7)#

The curve is passing through point #(6,-9)#

Then -

#y=c(x+3)(x-7)#

#-9 = c(6+3)(6-7)#
#-9=c(9)(-1)#
#-9=-9c#
#c=(-9)/(-9)=1#

Then the final equation is -

#y=1(x+3)(x-7)#

#y=(x+3)(x-7)#

We shall test it.

The Table for the above function.
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Plot all the points. Graph it.
Examine whether the curve passes through the critical points

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