Question

which of the following is an equation of the line tangent to the graph of $h(x)=x^{5}+3x^{2}+2x$ at the point where $x = -1$?
choose 1 answer:
a $y=-x + 1$
b $y=x - 1$
c $y=-x - 1$
d $y=x + 1$

Explanation:

Step1: Find the derivative of \(h(x)\)

Using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(h^\prime(x)=5x^{4}+6x + 2\).

Step2: Evaluate \(h(-1)\) and \(h^\prime(-1)\)

• For \(h(-1)\): \(h(-1)=(-1)^{5}+3(-1)^{2}+2(-1)=-1 + 3-2=0\).
• For \(h^\prime(-1)\): \(h^\prime(-1)=5(-1)^{4}+6(-1)+2=5 - 6 + 2=1\).

Step3: Use the point - slope form \(y - y_1=m(x - x_1)\)

Here \(x_1=-1,y_1 = 0,m = 1\). Substituting into \(y - y_1=m(x - x_1)\), we get \(y-0=1\times(x + 1)\), which simplifies to \(y=x + 1\).

Answer:

D. \(y = x + 1\)