Let f(x)= 49-x^2
The slope of the tangent line to the graph of f(x) at the point (-7,0) is:
The equation of the tangent line to the graph of f(x) at (7,0) isy=mx+b for
- PuzzlingLv 72 months ago
I answered a question very similar to this yesterday, but it seems to have been deleted. Also, you have (-7,0) at one place and then (7,0) at the other. I'll show you the method for the first point, but you can easily switch to x=7 to find the slope (m) and y-intercept (b) for the tangent line at the other point.
STEP 1 - Calculate the first derivative.
f'(x) = -2x
STEP 2 - Use that to find the slope at x=-7:
f'(-7) = -2(-7) = 14
m = 14
STEP 3 - Write the equation with the slope included.
y = 14x + b
STEP 4 - Plug in the point (-7,0)
0 = 14(-7) + b
0 = -98 + b
b = 98
y = 14x + 98
Check the graph of the function and the tangent line in the link below.Source(s): https://www.desmos.com/calculator/6xmlpy2uvv
- az_lenderLv 72 months ago
f'(x) = -2x,
so M at (-7,0) is 14.
The equation is y = 14x + B,
and the fact that (-7,0) is on the line shows that
B = 0 + 14(7) = 98.