Let f(x)=21−x^2
The slope of the tangent line to the graph of f(x) at the point (−4,5) is ______ .
The equation of the tangent line to the graph of f(x) at (−4,5) is y=mx+b for
m=______
and
b=_______.
f(x) = 21-x²
f'(x) = -2x
f'(-4) = 8
The line through (-4,5) with slope 8 is
(y-5)/(x+4) = 8
y = 8x +37
I think that's enough of these.
To find the slope of the tangent line to the graph of f(x) at the point (-4, 5), we need to take the derivative of f(x) and evaluate it at x = -4.
Step 1: Find the derivative of f(x)
f'(x) = d/dx (21 - x^2)
= -2x
Step 2: Evaluate the derivative at x = -4
f'(-4) = -2(-4)
= 8
Therefore, the slope of the tangent line to the graph of f(x) at the point (-4, 5) is 8.
Now, to find the equation of the tangent line, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point (-4, 5) and m is the slope.
Step 3: Substitute the values into the equation
y - 5 = 8(x - (-4))
Simplifying the equation further,
y - 5 = 8(x + 4)
y - 5 = 8x + 32
Comparing this equation with y = mx + b, we can see that the slope, m, is 8 and the y-intercept, b, is -37.
Therefore, the equation of the tangent line to the graph of f(x) at (-4, 5) is y = 8x - 37.
To find the slope of the tangent line to the graph of the function f(x) at a specific point, you need to take the derivative of the function and then substitute the x-coordinate of the point into the derivative.
Step 1: Find the derivative of f(x)
The derivative, denoted as f'(x) or dy/dx, represents the slope of the tangent line at any point on the graph. In this case, f(x) = 21 - x^2, so we need to find f'(x).
Using the power rule, the derivative of x^n is nx^(n-1), where n is a constant and x is the variable. Applying this rule to f(x) = 21 - x^2, differentiate each part of the function separately:
f'(x) = 0 - 2x
= -2x
Step 2: Substitute the x-coordinate into the derivative
Now that we have the derivative of f(x), we can find the slope of the tangent line at any specific point by substituting the x-coordinate of that point into the derivative.
To find the slope at the point (-4, 5), substitute x = -4 into f'(x):
m = -2(-4)
= 8
Therefore, the slope of the tangent line to the graph of f(x) at the point (-4, 5) is 8.
Step 3: Find the equation of the tangent line using the point-slope form
Once you have the slope of the tangent line, you can use the point-slope form of a linear equation to find the equation of the tangent line.
The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) represents the point on the line, and m is the slope of the line.
Using the point (-4, 5) and the slope m = 8, we can write the equation of the tangent line as:
y - 5 = 8(x - (-4))
Simplifying the equation further:
y - 5 = 8(x + 4)
y - 5 = 8x + 32
y = 8x + 37
Therefore, the equation of the tangent line to the graph of f(x) at the point (-4, 5) is y = 8x + 37.