find an equation of the tangent line to the graph of y=e^(2x-3) at point (3/2,1)
thanks!
y'(x) = 2e^(2x-3)
y'(3/2) = 2
so, you want the line through (3/2,1) with slope 2:
y-1 = 2(x-3/2)
Oh, finding the equation of a tangent line, huh? Well, I'm glad you're not asking me to find the equation of a clown car because that would involve a lot more honking!
Anyway, let's get back to math. To find the equation of a tangent line, we need to find the slope first. The slope of a tangent line to a function at a given point is equal to the derivative of the function evaluated at that point.
So, let's find the derivative of y = e^(2x-3). The derivative of e^(kx) is k * e^(kx), where k is a constant. In this case, k = 2. So, the derivative of y = e^(2x-3) is 2 * e^(2x-3).
Now, we can find the slope of the tangent line by plugging in x = 3/2 into the derivative. Slope = 2 * e^(2(3/2)-3) = 2 * e^0 = 2.
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 is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Plugging in the values, we get y - 1 = 2(x - 3/2).
And if we simplify it a bit, we get y - 1 = 2x - 3.
So, the equation of the tangent line to the graph of y = e^(2x-3) at point (3/2, 1) is y = 2x - 2.
I hope that made you smile!
To find the equation of the tangent line to the graph of y = e^(2x - 3) at the point (3/2, 1), you need to find the slope of the tangent line and use the point-slope form of a line.
Step 1: Find the derivative of the function y = e^(2x - 3) with respect to x.
dy/dx = d/dx(e^(2x - 3))
= e^(2x - 3)(d/dx(2x - 3))
= 2e^(2x - 3)
Step 2: Evaluate the derivative at x = 3/2 to find the slope of the tangent line at that point.
dy/dx = 2e^(2 * (3/2) - 3)
= 2e^0
= 2
The slope of the tangent line is 2.
Step 3: Use the point-slope form of a line: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
Plugging in the values, we have:
y - 1 = 2(x - 3/2)
Simplifying:
y - 1 = 2x - 3
y = 2x - 2
So, the equation of the tangent line to the graph of y = e^(2x - 3) at the point (3/2, 1) is y = 2x - 2.
To find the equation of the tangent line to the graph of a function at a given point, we need to calculate the slope of the tangent line as well as the coordinates of the point of tangency.
To find the slope of the tangent line, we need to find the derivative of the function at that point.
Let's start by finding the derivative of the function y = e^(2x - 3). The derivative of e^(kx) (where k is a constant) is ke^(kx).
Applying this rule, we get:
dy/dx = (d/dx)(e^(2x - 3)) = 2e^(2x - 3)
Now, we can find the slope of the tangent line by substituting the x-coordinate of the given point (3/2) into the derivative:
slope_m = dy/dx = 2e^(2(3/2) - 3) = 2e^(-3/2)
Next, we need to find the y-coordinate of the given point (3/2, 1). We can substitute this point into the original function to get the corresponding y-value:
y = e^(2x - 3)
1 = e^(2(3/2) - 3) = e^0 = 1
So, the y-coordinate of the given point is 1.
Now, we have the slope (m = 2e^(-3/2)) and the coordinates of the point of tangency (3/2, 1).
The equation of a line can be written in the form y = mx + b, where m is the slope and b is the y-intercept. We can substitute the coordinates of the point of tangency into this equation and solve for b.
Using the point-slope form, we have:
1 = 2e^(-3/2) * (3/2) + b
1 = 3e^(-3/2) + b
To solve for b, subtract 3e^(-3/2) from both sides:
b = 1 - 3e^(-3/2)
Thus, the equation of the tangent line to the graph of y = e^(2x - 3) at the point (3/2, 1) is:
y = 2e^(-3/2)x + (1 - 3e^(-3/2))