find the point on the graph of the given function at which the slope of the tangent line is the given slope
x^3+9x^2+33x+15 slope= 6
You should have properly written the function as
y = x^3 + 9x^2 + 33x + 15
or
f(x) = x^3 + 9x^2 + 33x + 15
then dy/dx = 3x^2 + 18x + 33
but you want that to be 6
3x^2 + 18x + 33 = 6
x^2 + 6x + 9 = 0
(x+3)^2 = 0
x+3 = 0
x = -3
then f(3) = -27 + 81 - 99 + 15 = -30
The point is (-3,-30)
To find the point on the graph of the given function where the slope of the tangent line is 6, we need to find the derivative of the function first.
Step 1: Find the derivative of the function
The derivative of a function gives us the slope of the tangent line at any point on the graph. For the given function f(x) = x^3 + 9x^2 + 33x + 15, we can use the power rule to find its derivative.
f'(x) = 3x^2 + 18x + 33
Step 2: Set up an equation with the given slope
We want to find the point where the slope of the tangent line is 6. So, we set up the equation:
f'(x) = 6
Step 3: Solve the equation
To solve the equation, we substitute f'(x) with its expression:
3x^2 + 18x + 33 = 6
Now, we can solve this quadratic equation. Rearranging the terms and setting the equation equal to zero, we have:
3x^2 + 18x + 27 = 0
Step 4: Solve the quadratic equation
There are a few methods to solve a quadratic equation. In this case, we can factor out a common factor:
3(x^2 + 6x + 9) = 0
Now, we have a perfect square trinomial within the parentheses:
3(x + 3)(x + 3) = 0
Simplifying further, we have:
(x + 3)^2 = 0
Taking the square root of both sides, we get:
x + 3 = 0
Solving for x, we find:
x = -3
Step 5: Find the y-coordinate
To find the y-coordinate corresponding to x = -3, we can substitute this value into the original function:
f(-3) = (-3)^3 + 9(-3)^2 + 33(-3) + 15
= -27 + 81 - 99 + 15
= -30
Therefore, the point on the graph of the given function where the slope of the tangent line is 6 is (-3, -30).