Determine the equation of the tangent line to the curve y = 2x²-1 at the point where x = -2
y' = slope = m = 4 x
where x = -2, m = 4(-2) = -8
y = -8 x + b
at x = -2, y = 2(4) - 1 = 7
so
7 = -8(-2) + b
7 = 16 + b
b = - 9
so
y = -8 x - 9
thank you
To determine the equation of the tangent line to the curve at the point where x = -2, we need to find the slope of the tangent line and the coordinates of the point on the curve.
Step 1: Find the coordinates of the point on the curve where x = -2.
To find the y-coordinate, substitute x = -2 into the equation:
y = 2(-2)² - 1
y = 2(4) - 1
y = 8 - 1
y = 7
So, the coordinates of the point on the curve where x = -2 are (-2, 7).
Step 2: Find the slope of the tangent line.
To find the slope of the tangent line, we need to find the derivative of the curve.
The derivative of y = 2x² - 1 can be found using the power rule for differentiation:
dy/dx = 2 * 2x = 4x.
Step 3: Substitute x = -2 into the derivative to find the slope at that point.
Slope = 4(-2) = -8.
So, the slope of the tangent line at the point (-2, 7) is -8.
Step 4: Write the equation of the tangent line using the point-slope form.
Using the point-slope form y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope:
y - 7 = -8(x - (-2))
y - 7 = -8(x + 2)
y - 7 = -8x - 16
y = -8x - 9
Therefore, the equation of the tangent line to the curve y = 2x² - 1 at the point where x = -2 is y = -8x - 9.
To determine the equation of the tangent line to a curve, we need to find the slope of the curve at the given point. Once we have the slope, we can use the point-slope form of a line to write the equation.
Step 1: Find the derivative of the function y = 2x² - 1. The derivative gives us the slope of the tangent line at any given point on the curve.
The derivative of 2x² is 4x, and the derivative of -1 is 0. So, the derivative of y = 2x² - 1 is dy/dx = 4x.
Step 2: Evaluate the derivative at x = -2 to find the slope at that point.
Plug in x = -2 into the derivative equation: dy/dx = 4x. We get dy/dx = 4(-2) = -8.
Therefore, the slope of the tangent line to the curve y = 2x² - 1 at the point where x = -2 is -8.
Step 3: Use the point-slope form of a line to write the equation of the tangent line.
The point-slope form of a line is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, m is the slope of the line.
We know the slope, m = -8, and the point (x₁, y₁) is (-2, y), where y is the y-coordinate at x = -2 on the curve.
Plug in the values:
y - y₁ = m(x - x₁)
y - y₁ = -8(x - (-2))
y - y₁ = -8(x + 2)
Since the tangent line passes through the point (-2, y) on the curve, we can substitute x = -2 into the original equation to find y.
y = 2x² - 1
y = 2(-2)² - 1
y = 2(4) - 1
y = 8 - 1
y = 7
Therefore, the point of tangency is (-2, 7).
Now we have all the information to write the equation of the tangent line:
y - y₁ = -8(x + 2)
y - 7 = -8(x + 2)
This is the equation of the tangent line to the curve y = 2x² - 1 at the point where x = -2.