Find the equation of the line tangent to the graph of y=4 e^x at x=2.
I always get confused with tangent lines
I worked this out to be
4e^x(x-2)+4e^2 but the program i have to submit it to says its wrong.
The equation of the tangent to the graph
y=f(x) at x=a:
y=f(a)+f'(a)(x-a)
(4e^x)'=4e^x
y=4e^2+4e^2(x-2)
y=4e^2*x-4e^2
To find the equation of the line tangent to the graph of y = 4e^x at x = 2, we can use the derivative of the function.
First, let's find the derivative of y = 4e^x with respect to x.
The derivative of y = 4e^x can be found using the chain rule, which states that if we have a composite function such as f(g(x)), the derivative is given by f'(g(x)) * g'(x).
In this case, f(x) = 4e^x and g(x) = x. So the derivative is:
dy/dx = f'(g(x)) * g'(x)
= (4e^x)' * (x)'
The derivative of e^x is just e^x. The derivative of x is 1. So we have:
dy/dx = 4e^x * 1
= 4e^x
Now, we have the derivative of the function: dy/dx = 4e^x.
To find the slope of the tangent line at x = 2, we substitute x = 2 into the derivative:
slope = 4e^2
The equation for a line with slope m and passing through the point (x1, y1) is given by the point-slope form:
y - y1 = m(x - x1)
Substituting the values y1 = 4e^2 and x1 = 2, and using the slope we found earlier:
y - (4e^2) = (4e^2)(x - 2)
Expanding this equation:
y - 4e^2 = 4e^2x - 8e^2
Adding 4e^2 to both sides:
y = 4e^2x - 4e^2 + 4e^2
Simplifying:
y = 4e^2x
Therefore, the equation of the line tangent to the graph of y = 4e^x at x = 2 is y = 4e^2x.
To find the equation of the tangent line to the graph of y = 4e^x at x = 2, you'll need to find both the slope and the y-intercept of the tangent line.
First, let's find the derivative of the function y = 4e^x. The derivative of e^x is e^x, so using the chain rule, the derivative of y = 4e^x is:
dy/dx = 4e^x
Next, evaluate the derivative at x = 2 to find the slope of the tangent line:
dy/dx = 4e^2
Now that we have the slope, we can find the y-intercept by substituting the coordinates (2, y) into the equation y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point (2, y) on the curve.
Using the point-slope form, we have:
y - y1 = (4e^2)(x - 2)
Simplifying further, we have:
y - y1 = 4e^2x - 8e^2
Now, let's solve for y:
y = 4e^2x - 8e^2 + y1
Since we're looking for the equation of the tangent line, we don't have the exact value for y1. However, we can still represent it as a constant:
y = 4e^2x - 8e^2 + C
So, the equation of the tangent line to the graph of y = 4e^x at x = 2 is y = 4e^2x - 8e^2 + C, where C is a constant value.