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The graph of a certain quadratic \(y = ax^2 + bx + c\)  is a parabola with vertex \( (-4,0)\)  which passes through the point (1,-75). What is the value of a?

tertre  Mar 12, 2017

Best Answer 

 #1
avatar
+5

Im not sure if this is right or not but I had a go:

 

y=ax2+bx+c

 

use vertex/turning point form

y=a(x-h)2+k

sub in vertex

y=a(x+4)2

sub in point values and solve for a

-75=a(1+4)2

-75=25a

a=-3

 

Hope this helps.

Guest Mar 12, 2017
edited by Guest  Mar 12, 2017
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2+0 Answers

 #1
avatar
+5
Best Answer

Im not sure if this is right or not but I had a go:

 

y=ax2+bx+c

 

use vertex/turning point form

y=a(x-h)2+k

sub in vertex

y=a(x+4)2

sub in point values and solve for a

-75=a(1+4)2

-75=25a

a=-3

 

Hope this helps.

Guest Mar 12, 2017
edited by Guest  Mar 12, 2017
 #2
avatar+1148 
+5

That is correct!

tertre  Mar 12, 2017

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