write an equation of the line tangent to the graph of $f(x)=2x^{2}+7x - 9$ at the point where $x=-3$.

Explanation:

Step1: Find the function value at (x = - 3)

Substitute (x=-3) into (f(x)=2x^{2}+7x - 9). (f(-3)=2\times(-3)^{2}+7\times(-3)-9=2\times9-21 - 9=18-21 - 9=-12)

Step2: Find the derivative of the function

Use the power rule ((x^{n})^\prime=nx^{n - 1}). (f^\prime(x)=(2x^{2}+7x - 9)^\prime=(2x^{2})^\prime+(7x)^\prime-(9)^\prime=4x + 7)

Step3: Find the slope of the tangent line at (x=-3)

Substitute (x = - 3) into (f^\prime(x)). (f^\prime(-3)=4\times(-3)+7=-12 + 7=-5)

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (m=-5), (x_{1}=-3), (y_{1}=-12). (y-(-12)=-5(x-(-3))) (y + 12=-5(x + 3)) Expand: (y+12=-5x-15) (y=-5x-27)

Answer:

(y=-5x - 27)