From first principles (ie using the tangent slope method), find the slope of the following curves at the given value of x.
f(x)=2x^2−6x at x = 3
f(x)=2x^2-6x
f(x+h)= 2(x+h)^2-6(x+h)
=2x^2+4xh+2h^2-6x-6h
lim h-->0 f(x+h)-f(x)/h
=2x^2+4xh+2h^2-6x-6h - (2x^2-6x)
=4xh+2h^2-6h/h
=h(4x+2h-6)/h
=lim h -->0 = 4x+2h-6
4x+2(0)-6
=4x-6
at x=3
f(3)=4x-6
=4(3)-6
=6
Im I doing this correct?
you are right on.
Are you sure? Someone told me I'm doing this wrong
Yes, you have correctly calculated the slope of the curve f(x) = 2x^2 - 6x at the given value of x, which is x = 3.
To find the slope using the tangent slope method, you first find the derivative of the function f(x) with respect to x. In this case, the derivative of f(x) = 2x^2 - 6x is f'(x) = 4x - 6.
Then, you substitute the value of x (which is 3 in this case) into the derivative to get the slope at that point:
f'(3) = 4(3) - 6
= 12 - 6
= 6
So, the slope of the curve f(x) = 2x^2 - 6x at x = 3 is 6.