Question

y = (x² + 4)(x⁵ - 3x) at x = 1
find the equation of the line tangent to
y + 10 = 6(x - 1)
y - 10 = 6(x - 1)
y + 10 = -6(x - 1)
y - 10 = -6(x + 1)

Explanation:

Step1: Find the value of \(y\) at \(x = 1\)

Substitute \(x=1\) into \(y=(x^{2}+4)(x^{5}-3x)\):
\(y=(1^{2}+4)(1^{5}-3\times1)=(1 + 4)(1-3)=5\times(-2)=-10\)

Step2: Differentiate \(y=(x^{2}+4)(x^{5}-3x)\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\)

Let \(u=x^{2}+4\), then \(u^\prime = 2x\); let \(v=x^{5}-3x\), then \(v^\prime=5x^{4}-3\)
\(y^\prime=(2x)(x^{5}-3x)+(x^{2}+4)(5x^{4}-3)\)

Step3: Find the slope of the tangent line at \(x = 1\)

Substitute \(x = 1\) into \(y^\prime\):
\(y^\prime|_{x = 1}=(2\times1)(1^{5}-3\times1)+(1^{2}+4)(5\times1^{4}-3)\)
\(=2\times(1 - 3)+5\times(5 - 3)\)
\(=2\times(-2)+5\times2\)
\(=-4 + 10=6\)

Step4: Use the point - slope form \(y - y_{0}=m(x - x_{0})\)

Here \(x_{0}=1\), \(y_{0}=-10\), \(m = 6\)
\(y-(-10)=6(x - 1)\), which simplifies to \(y + 10=6(x - 1)\)

Answer:

A. \(y + 10=6(x - 1)\)