Find equations of the tangent line and normal line to the curve at the given point.
y = x^4 + 2e^x, (0,2)
Never mind, I got the answer :)
To find the equations of the tangent line and normal line to the curve at a given point, we first need to find the slope of the tangent line at that point.
1. Find the derivative of the given function:
To find the slope of the tangent line, we need to find the derivative of the function y = x^4 + 2e^x.
Taking the derivative of each term separately, we get:
dy/dx = d/dx(x^4) + d/dx(2e^x)
= 4x^3 + 2e^x
2. Calculate the slope of the tangent line:
Substitute the x-coordinate of the given point (0) into the derivative we found in step 1:
slope = (4 * 0^3) + 2e^0
= 0 + 2
= 2
3. Find the equation of the tangent line:
Now that we have the slope of the tangent line, we can use the point-slope form of a line to find its equation.
The point-slope form is given by: y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope.
Plugging in the values, we have:
y - 2 = 2(x - 0)
y - 2 = 2x
y = 2x + 2
4. Find the equation of the normal line:
The normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the tangent line's slope.
The negative reciprocal of 2 is -1/2.
Again, use the point-slope form to find the equation of the normal line:
y - 2 = (-1/2)(x - 0)
y - 2 = -1/2x
y = -1/2x + 2
Therefore, the equation of the tangent line to the curve y = x^4 + 2e^x at the point (0, 2) is y = 2x + 2, and the equation of the normal line is y = -1/2x + 2.