Solution:
Given:
[tex]f(x)=2x^4-5[/tex]We are required to find the equation of the tangent line to the graph of the function at the given point (-1,-3)
Firstly, determine the slope of the curve at the given point
[tex]\begin{gathered} f(x)=2x^4-5 \\ f^1(x)=8x^3 \\ At\text{ point \lparen-1,-3\rparen, } \\ Subsitute\text{ x=-1 into f}^1(x) \\ slope,m=8(-1)^3 \\ m=-8 \end{gathered}[/tex]The tangent to the curve at (-1, -3) has the slope -8
[tex]\begin{gathered} Recall\text{ that } \\ y-y_1=m(x-x_1) \\ m=-8,\text{ x}_1=-1,\text{ y}_1=-3 \\ y-(-3)=-8(x-(-1)) \\ y+3=-8(x+1) \\ y+3=-8x-8 \\ y=-8x-8-3 \\ y=-8x-11 \end{gathered}[/tex]The picture below show the illustration of the situation
The answer is y = -8x - 11
