Question

let (f(x)=x^{3}-6x + 10). then the equation of the tangent line to the graph of (f(x)) at the point ((-4,-30)) is given by (y = mx + b) for (m=) and (b=)

Explanation:

Step1: Find the derivative of f(x)

The derivative of $f(x)=x^{3}-6x + 10$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=3x^{2}-6$.

Step2: Calculate the slope m

Substitute $x=-4$ into $f'(x)$. So $m = f'(-4)=3\times(-4)^{2}-6=3\times16 - 6=48 - 6=42$.

Step3: Find the y - intercept b

We know the line $y = mx + b$ passes through the point $(-4,-30)$ and $m = 42$. Substitute $x=-4$, $y=-30$ and $m = 42$ into $y=mx + b$. We get $-30=42\times(-4)+b$. Then $-30=-168 + b$. Solving for $b$ gives $b=-30 + 168=138$.

Answer:

$m = 42$
$b = 138$