Determine without graphing, whether the given quadratic function has a max value or min value and then find the value.
F(x)=2x^2+16x-1
since you have 2x^2, the parabola opens upward
So, it must have a minimum, which occurs at x = -16/4
f(x)=2(x^2+8x) -1
= 2(x^2+4x+4) -1-8
= 2(x+2)^2 -9
sp as x gets large, either negative or positive, f(x) increases, which means at x=-2 it is a minimum.
To determine whether the given quadratic function has a maximum value or minimum value, we can look at the coefficient of the x^2 term.
In the quadratic function F(x) = 2x^2 + 16x - 1, the coefficient of the x^2 term is positive (2), which means it opens upwards.
When a quadratic function opens upwards, it has a minimum value. Conversely, if the coefficient of the x^2 term were negative, it would open downwards and have a maximum value.
To find the value of the minimum point, we need to use the formula for the x-coordinate of the vertex:
x = -b / (2a)
Where a, b, and c are the coefficients of the quadratic function in the form ax^2 + bx + c.
In this case, a = 2 and b = 16:
x = -16 / (2 * 2) = -16 / 4 = -4
So, the x-coordinate of the minimum point is -4.
To find the corresponding y-coordinate, we substitute this value back into the original function:
F(x) = 2x^2 + 16x - 1
F(-4) = 2(-4)^2 + 16(-4) - 1
= 32 - 64 - 1
= -33
Therefore, the quadratic function F(x) = 2x^2 + 16x - 1 has a minimum value of -33, which occurs at x = -4.