Question

find the slope and the equation of the tangent line to the graph of the function at the given value of x.
f(x)=x^4 - 5x^2 + 4; x = 1

Explanation:

Step1: Find the derivative of the function

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=x^{4}-5x^{2}+4$, then $f'(x)=4x^{3}-10x$.

Step2: Calculate the slope of the tangent line

Substitute $x = 1$ into $f'(x)$. So $m=f'(1)=4(1)^{3}-10(1)=4 - 10=-6$.

Step3: Find the y - coordinate of the point on the curve

Substitute $x = 1$ into $f(x)$. $y=f(1)=1^{4}-5(1)^{2}+4=1 - 5 + 4=0$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,0)$ and $m=-6$. So $y-0=-6(x - 1)$, which simplifies to $y=-6x + 6$.

Answer:

Slope: $-6$; Equation of the tangent line: $y=-6x + 6$