find an equation of the tangent line to the parabola at the given point..show work
y=(-2)x^2
(-1,-2)
I assume you know enought calc to take the derivative.
y'=slope=-4x=4
y=4x+b
solve for b a the point, and you have the line
dont know how to find the derivitive
To find the equation of the tangent line to the parabola at the given point, we need to find the slope of the tangent line first.
The slope of the tangent line at a given point on a curve is equal to the derivative of the equation of the curve at that point.
Given that the equation of the parabola is y = -2x^2, we can find its derivative using the power rule.
Take the derivative of -2x^2:
dy/dx = -2 * 2x = -4x
Now, substitute the x-coordinate of the given point (-1, -2) into the derivative equation to find the slope at that point:
slope = -4 * (-1) = 4
So, the slope of the tangent line at the point (-1, -2) is 4.
Now that we have the slope, we can use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by: y - y₁ = m(x - x₁), where (x₁, y₁) is the given point on the line, and m is the slope of the line.
Substituting the values (-1, -2) and m = 4 into the equation, we get:
y - (-2) = 4(x - (-1))
y + 2 = 4(x + 1)
Simplifying the equation, we have:
y + 2 = 4x + 4
y = 4x + 2 - 4
y = 4x - 2
Therefore, the equation of the tangent line to the parabola y = -2x^2 at the point (-1, -2) is y = 4x - 2.