Let f(x)=3x^4-2x^2+3
a) find f'(x)
Answer: f'(x)=12x^3-4x
b) find the slope of the line tangent to the graph of f at x=3.
Answer:312
So I got those firs two right but what I need help with is part C which is
c) find an equation of the line tangent to the graph of f at x=3
what does ^ mean
to the power of, for example x squared or x cubed
surely by now you know that the line through (h,k) with slope m is
y-k = m(x-h)
you have h=3,m=312, and k=f(3)
plug and chug.
To find an equation of the line tangent to the graph of f at x=3, you can use the point-slope form of a linear equation. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line, and m is the slope of the line.
We already know the slope from part b), which is 312. And since we want the line tangent to the graph at x=3, we can substitute x1 = 3.
Now let's find the corresponding y-coordinate on the graph of f(x) at x=3. Substitute x=3 into the original function f(x):
f(3) = 3(3)^4 - 2(3)^2 + 3
= 81 - 18 + 3
= 66
So we have the point (3, 66) on the line.
Now we can substitute the values into the point-slope form to get the equation of the line:
y - 66 = 312(x - 3)
Simplifying:
y - 66 = 312x - 936
y = 312x - 870
Therefore, an equation of the line tangent to the graph of f at x=3 is y = 312x - 870.