2 years ago

H(x) = ax^2 +1 Find the value of a when h(-2) = 21.

Mathematics

1 answer:

blagie [28]2 years ago

4 0

Answer:

$a = 5$

Step-by-step explanation:

Solving $h(-2) = 21$ :

$h(-2) = 21 \\ a(-2)^2 +1 = 21 \\ a(4) +1 = 21 \\ 4a +1 = 21 \\ 4a +1 -1 = 21 -1 \\ 4a = 20 \\ \frac{4a}{4} = \frac{20}{4} \\ a = 5$

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