find the points where the graph of the function have horizontal tangents
f(x) = 4x^2 + 6x - 5
To find the points where the graph of the function has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.
Let's find the derivative of f(x) first:
f(x) = 4x^2 + 6x - 5
f'(x) = 8x + 6
Now, let's set f'(x) = 0 and solve for x:
8x + 6 = 0
8x = -6
x = -6/8
x = -3/4
So, the point where the graph of the function has a horizontal tangent is (-3/4, f(-3/4)).
To find the y-coordinate of this point, let's substitute x = -3/4 into the original function:
f(-3/4) = 4(-3/4)^2 + 6(-3/4) - 5
f(-3/4) = 4(9/16) - 18/4 - 5
f(-3/4) = 9/4 - 9/2 - 5
f(-3/4) = 9/4 - 18/4 - 20/4
f(-3/4) = (9 - 18 - 20)/4
f(-3/4) = -29/4
Therefore, the point where the graph of the function has a horizontal tangent is (-3/4, -29/4).