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# Graphing question. Halp

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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

#1
+5788
+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
+5788
+3

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

Rom Aug 12, 2019
#3
+771
+3

very good Rom

# 👍👍👍👍

here scince i cant give you more points theres four extra

travisio  Aug 12, 2019
#2
+1

Thank you SO MUCH.

Aug 12, 2019