Find the vertex of F(x) by using the formula for the x-coordinate of the vertex.
F(x)= 4x^2 - 3x - 6
I don't exactly understand this question. could someone clear it up and help me solve it? Thanks!!
factor it:
F(x)=4(x^2-3/4x ) -6
F(x)=4(x^2-3/4 x + 9/64) - 6 - 4*9/64
= 4(x-3/8)^2 -( )
Well,it is obvious that x=3/8 is the x vertix. Putting that x in, then y= -6-36/64
check my thinking.
Finding the vertex of a quadratic function can be done using the formula for the x-coordinate of the vertex, which is given by x=-b/2a. The general form of a quadratic function is F(x) = ax^2 + bx + c, where a, b, and c are coefficients.
In this case, the quadratic function is F(x) = 4x^2 - 3x - 6. Comparing this to the general form, we can see that a = 4, b = -3, and c = -6.
To find the x-coordinate of the vertex, we can substitute these values into the formula x = -b/2a.
x = -(-3) / (2 * 4)
x = 3 / 8
So, the x-coordinate of the vertex is 3/8.
Now, to find the y-coordinate of the vertex, we can substitute this x-coordinate back into the original function.
F(3/8) = 4(3/8)^2 - 3(3/8) - 6
=(9/8) - (9/8) - 6
=(9/8) - (9/8) - 6*(8/8)
= (9 - 9 - 48) / 8
= -48/8
= -6
Therefore, the vertex of the function F(x) = 4x^2 - 3x - 6 is (3/8, -6).