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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 \(a\)?

 

Parabolas are hard for me. Help. Plz

 Aug 12, 2019

Best Answer 

 #1
avatar+6046 
+3

\(\text{Writing this in vertex form we have}\\ y = \alpha(x+4)^2 + 0 = \alpha(x+4)^2\\ \text{using the point we're given}\\ -75 = \alpha(1+4)^2 = 25\alpha\\ \alpha = -3\\~\\ y = -3(x+4)^2 = -3(x^2 + 8x+16)\\ y = -3x^2 - 24x - 48\\ a = -3\)

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 Aug 12, 2019
 #1
avatar+6046 
+3
Best Answer

\(\text{Writing this in vertex form we have}\\ y = \alpha(x+4)^2 + 0 = \alpha(x+4)^2\\ \text{using the point we're given}\\ -75 = \alpha(1+4)^2 = 25\alpha\\ \alpha = -3\\~\\ y = -3(x+4)^2 = -3(x^2 + 8x+16)\\ y = -3x^2 - 24x - 48\\ a = -3\)

Rom Aug 12, 2019
 #3
avatar+792 
+3

very good Rom

👍👍👍👍

here scince i cant give you more points theres four extra

travisio  Aug 12, 2019
 #2
avatar
+1

Thank you SO MUCH.

 Aug 12, 2019

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