At what points on the graph of f(x)=6x^2+4x-9 is the slope of the tangent line -2?
dy/dx=-2=6x^2+4x-9
6x^2+4x-7=0
x=(-4+-sqrt(16+4*6*7))/12
x=(-4+-13.6)/12
those are the two points. Check my math with a calc, I did it in my head.
Bob forgot to find the derivative first
dy/dx = 12x + 4
12x+4 = -2
12x = -8
x = -8/12 = -2/3
f(-2/3) = 6(4/9) + 4(-2/3) - 9 = -25/3
We have a parabola shown below with a tangent at (-2/3 , -25/3) having a slope of -2
http://www.wolframalpha.com/input/?i=plot+y%3D6x%5E2%2B4x-9
To find the points on the graph of f(x)=6x^2+4x-9 where the slope of the tangent line is -2, we need to find the values of x that satisfy dy/dx=-2.
The derivative of f(x) can be found by differentiating each term separately:
f'(x) = d/dx (6x^2) + d/dx (4x) + d/dx (-9)
Simplifying this expression, we get:
f'(x) = 12x + 4
Now, we need to set f'(x) equal to -2 and solve for x:
12x + 4 = -2
Subtracting 4 from both sides:
12x = -6
Dividing both sides by 12:
x = -6/12
Simplifying, we get:
x = -1/2
Therefore, the x-value of the point(s) on the graph of f(x)=6x^2+4x-9 where the slope of the tangent line is -2 is -1/2.
To find the corresponding y-value(s), we can substitute x=-1/2 into the original function f(x):
f(-1/2) = 6(-1/2)^2 + 4(-1/2) - 9
Simplifying, we get:
f(-1/2) = 6(1/4) - 2 - 9
f(-1/2) = 3/2 - 11/2
f(-1/2) = -8/2
f(-1/2) = -4
So, the point on the graph of f(x) where the slope of the tangent line is -2 is (-1/2, -4).
To find the points on the graph of the function where the slope of the tangent line is -2, we first need to find the derivative of the function. The derivative gives us the slope of the tangent line at any given point on the graph.
The derivative of f(x) = 6x^2 + 4x - 9 can be found by taking the derivative of each term separately. The derivative of 6x^2 is 12x, the derivative of 4x is 4, and the derivative of -9 is 0. So, the derivative of f(x) is:
f'(x) = 12x + 4
Now, we need to solve the equation f'(x) = -2 to find the x-values where the slope of the tangent line is -2:
12x + 4 = -2
Subtracting 4 from both sides, we have:
12x = -6
Dividing both sides by 12, we get:
x = -6/12
Simplifying further, we have:
x = -1/2
So, the x-coordinate of the point on the graph where the slope of the tangent line is -2 is -1/2.
To find the corresponding y-coordinate, we can substitute this value of x back into the original function:
f(-1/2) = 6(-1/2)^2 + 4(-1/2) - 9
Simplifying further:
f(-1/2) = 6(1/4) - 2 - 9
f(-1/2) = 3/2 - 2 - 9
f(-1/2) = 3/2 - 18/2 - 9
f(-1/2) = -15/2 - 9
f(-1/2) = -15/2 - 18/2
f(-1/2) = -33/2
So, the point on the graph where the slope of the tangent line is -2 is (-1/2, -33/2).