Question

write an equation of the line tangent to the graph of (g(x)=x^{2}+5x - 3) at the point where (x=-3).

Explanation:

Step1: Find the derivative of $g(x)$

The derivative of $g(x)=x^{2}+5x - 3$ using the power - rule $(x^n)'=nx^{n - 1}$ is $g'(x)=2x + 5$.

Step2: Find the slope of the tangent line at $x=-3$

Substitute $x = - 3$ into $g'(x)$. So $m=g'(-3)=2(-3)+5=-6 + 5=-1$.

Step3: Find the y - coordinate of the point on the curve at $x=-3$

Substitute $x=-3$ into $g(x)$. $g(-3)=(-3)^{2}+5(-3)-3=9-15 - 3=-9$.

Step4: Use the point - slope form of a line $y - y_1=m(x - x_1)$

Here $x_1=-3$, $y_1=-9$ and $m=-1$. So $y+9=-1(x + 3)$.

Step5: Simplify the equation

$y+9=-x - 3$, which simplifies to $y=-x-12$.

Answer:

$y=-x - 12$