Find the equation of the tangent to y=x^2+6x+9 which has a slope of 6.
the slope of the tangent is the 1st derivative
2x + 6 = 6 ... x = 0 ... y = 9
the tangent line goes through (0,9) and has a slope of 6
using point-slope ... y - 9 = 6x
To find the tangent line with a slope of 6 to the curve given by y = x^2 + 6x + 9, we first need to find the derivative of the curve.
Step 1: Find the derivative of y = x^2 + 6x + 9 with respect to x.
The derivative of x^2 is 2x, and the derivative of 6x is 6. Since the derivative of a constant is zero, the derivative of 9 is zero.
Therefore, the derivative of y = x^2 + 6x + 9 is dy/dx = 2x + 6.
Step 2: Set the derivative equal to the slope.
Given that the slope of the tangent line is 6, we set the derivative equal to 6:
2x + 6 = 6
Step 3: Solve for x.
Subtracting 6 from both sides, we get:
2x = 0
Dividing both sides by 2:
x = 0
Step 4: Find the corresponding y-coordinate.
To find the y-coordinate, substitute the value of x into the original equation y = x^2 + 6x + 9:
y = (0)^2 + 6(0) + 9
= 0 + 0 + 9
= 9
Therefore, the point of tangency is (0, 9).
Step 5: Write the equation of the tangent line.
To find the equation of the tangent line, we need the slope (m) and a point (x1, y1) on the line.
We have the slope as 6, and the point of tangency is (0, 9).
Using the point-slope form of the equation of a line, y - y1 = m(x - x1), we substitute in the values:
y - 9 = 6(x - 0)
Simplifying:
y - 9 = 6x
y = 6x + 9
Therefore, the equation of the tangent line to y = x^2 + 6x + 9 with a slope of 6 is y = 6x + 9.
To find the equation of the tangent to the given curve, we need to find the point on the curve where the tangent line intersects and its slope.
Step 1: Find the derivative of the given function to get the slope of the tangent line at any point on the curve.
The derivative of y = x^2 + 6x + 9 with respect to x is:
dy/dx = 2x + 6
Step 2: Set the derivative equal to the given slope to find the x-coordinate of the point where the tangent line intersects.
We are given that the slope of the tangent line is 6. So, we have:
2x + 6 = 6
Step 3: Solve the equation from Step 2 to find the x-coordinate.
Subtracting 6 from both sides:
2x = 0
Dividing both sides by 2:
x = 0
Step 4: Substitute the x-coordinate into the original equation to find the y-coordinate.
Using the equation y = x^2 + 6x + 9, we substitute x = 0 to get:
y = 0^2 + 6(0) + 9
y = 9
So, the point of intersection of the tangent line and the curve is (0, 9).
Step 5: Use the point-slope form to write the equation of the tangent line using the point of intersection and given slope.
The point-slope form is:
y - y1 = m(x - x1)
Substituting the values we found, we get:
y - 9 = 6(x - 0)
Simplifying:
y - 9 = 6x
Rearranging to the standard form:
6x - y + 9 = 0
Therefore, the equation of the tangent line to the curve y = x^2 + 6x + 9 with a slope of 6 is 6x - y + 9 = 0.