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given h(x), evaluate h(-8). h(x)=\frac{x^{2}+3x}{4x + 27} h(-8)=?

Question

given h(x), evaluate h(-8). h(x)=\frac{x^{2}+3x}{4x + 27} h(-8)=?

Explanation:

Step1: Substitute x = -8

Substitute x=-8 into h(x)=$\frac{x^{2}+3x}{4x + 27}$.
$h(-8)=\frac{(-8)^{2}+3\times(-8)}{4\times(-8)+27}$

Step2: Calculate numerator

Calculate $(-8)^{2}+3\times(-8)$.
$(-8)^{2}+3\times(-8)=64 - 24=40$

Step3: Calculate denominator

Calculate $4\times(-8)+27$.
$4\times(-8)+27=-32 + 27=-5$

Step4: Calculate h(-8)

$h(-8)=\frac{40}{-5}=-8$

Answer:

-8