a.Find the slope of the tangent line to the curve y=9-2x^2 at the point (2,1)

b.Find an equation of this tangent line.

as you know, the slope of the tangent line is just the derivative. So,

y'(x) = -4x
giving us
y'(2) = -8

and now we have a point and a slope, so the line is

y-1 = -8(x-2)

To find the slope of the tangent line to the curve y = 9 - 2x^2 at the point (2,1), we can differentiate the equation with respect to x.

a. Differentiating y = 9 - 2x^2:
dy/dx = -4x

Next, substitute x = 2 into the derivative:

dy/dx = -4(2)
= -8

Therefore, the slope of the tangent line to the curve at the point (2,1) is -8.

b. To find the equation of the tangent line, we can use the point-slope form of a line. The equation can be written as:

y - y₁ = m(x - x₁)

Substituting the values:

y - 1 = -8(x - 2)

Expanding:

y - 1 = -8x + 16

Rearranging the equation:

y = -8x + 17

So the equation of the tangent line to the curve at the point (2,1) is y = -8x + 17.

a. To find the slope of the tangent line to the curve y = 9 - 2x^2 at the point (2,1), we need to find the derivative of the function at that point. The derivative gives the rate of change of the function at a specific point.

First, let's find the derivative of the function y = 9 - 2x^2. To do this, we'll differentiate each term separately. The derivative of the constant term 9 is 0 since it doesn't change with respect to x. The derivative of -2x^2 is -4x because we take the power (2) and multiply it with the coefficient (-2), and then subtract 1 from the power.

So, the derivative of y with respect to x is dy/dx = 0 - 4x = -4x.

Now, let's find the slope of the tangent line at the point (2,1) by substituting x = 2 into the derivative:

slope = dy/dx |(x=2) = -4(2) = -8.

Therefore, the slope of the tangent line to the curve y = 9 - 2x^2 at the point (2,1) is -8.

b. To find the equation of the tangent line, we'll use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.

We already know the slope, which is -8, and we have the point (2,1).

Using the point-slope form, we can substitute the values into the equation:

y - 1 = -8(x - 2).

To simplify, distribute -8 to (x - 2):

y - 1 = -8x + 16.

Now, isolate y by adding 1 to both sides:

y = -8x + 17.

Therefore, the equation of the tangent line to the curve y = 9 - 2x^2 at the point (2,1) is y = -8x + 17.