Find the equation of the tangent line to f(x)=sqrtx^2+6x at x=2, give your answer in slope-intercept form
can't tell if you mean
f(x) = √(x^2) + 6x
or
f(x) = √(x^2 + 6x)
I will assume the latter
f'(x) = (1/2)(2x)/√(x^2 + 6x)
f'(2) = 2/√16 = 1/2
also f(2) = .....
So now you have the slope and a point, you must be able to finish it
I meant f(x)sqrt(x^2+6x), thank you!
Am I doing this correct?
y-y1=m(x-x1)
y-1/2=-1/2(x-2)
y-1/2=-1/2x+1
add 1/2 to both sides
y=-1/2x+1.5
Your y value of the point is not correct
f(2) = √(2^2 + 6(2)) = √ 16 = 4
so the point is (2,4) and the slope is 1/2
y-4 = (1/2)(x-2)
y = (1/2)x - 1 + 3
y = (1/2)x + 2
To find the equation of the tangent line to the function f(x) = sqrt(x^2 + 6x) at the point x = 2, we need to determine both the slope and the y-intercept of the tangent line.
Step 1: Find the derivative of f(x)
The derivative of f(x) can be found using the power rule. Let's differentiate the function f(x) = sqrt(x^2 + 6x) term by term:
f'(x) = (1/2)*(x^2 + 6x)^(-1/2) * (2x + 6)
Simplifying further, we get:
f'(x) = (x + 3) / sqrt(x^2 + 6x)
Step 2: Find the slope of the tangent line
To find the slope of the tangent line at x = 2, substitute this value into the derivative:
f'(2) = (2 + 3) / sqrt(2^2 + 6*2)
= 5 / sqrt(16)
= 5 / 4
Therefore, the slope of the tangent line is 5/4.
Step 3: Find the y-coordinate at the point of tangency
To find the y-coordinate at x = 2, substitute this value into the original function:
f(2) = sqrt(2^2 + 6*2)
= sqrt(4 + 12)
= sqrt(16)
= 4
So, the y-coordinate at the point of tangency is 4.
Step 4: Write the equation of the tangent line in slope-intercept form
The equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. We already found the slope (m = 5/4), and we have the point (2, 4) as the point of tangency.
Using the point-slope form, we can rewrite the equation as:
y - 4 = (5/4) * (x - 2)
Simplifying further:
y - 4 = (5/4)x - 5/2
Now, solve for y by adding 4 to both sides of the equation:
y = (5/4)x - 5/2 + 4
Simplifying the expression:
y = (5/4)x + 3/2
Therefore, the equation of the tangent line to f(x) = sqrt(x^2 + 6x) at x = 2 is y = (5/4)x + 3/2.