min/max point
minimum or maximum
f(x)=2x^2+8x-2
a=2
x=(-b)/(2a)=(-8)/(2)(2)=-4
x=-4
y=2(-4)^2+8(-4)-2
2(16)-32-2
32-32-2
y=-2
(-4,-2)
axis of symmetry is x=-4
range[-2,oo]
minimum value -2
I think you messed up when trying to find the axis of symmetry. There's a small calculation error there and it seems to have thrown off all your answers. However, your work overall is correct.
when you used the calculator to find (-8)/(2)(2) you did not consider the order of operations you must multiply 2 by 2 before you divide -8 by the answer, otherwise your method is correct
My range changes to -10 and the axis is x=-2, and min value is -10?
Yes.
To find the minimum or maximum point of a quadratic function, you can start by determining the vertex of the parabola.
For the given quadratic function f(x) = 2x^2 + 8x - 2, the formula to find the x-coordinate of the vertex is x = (-b) / (2a), where a and b are the coefficients of the quadratic term and linear term, respectively.
In this case, a = 2 and b = 8, so x = (-8) / (2 * 2) = -4. The axis of symmetry is given by the value of x, so x = -4.
To find the y-coordinate of the vertex, substitute the value of x back into the original function. For x = -4:
f(-4) = 2(-4)^2 + 8(-4) - 2
f(-4) = 2(16) - 32 - 2
f(-4) = 32 - 32 - 2
f(-4) = -2
Therefore, the minimum or maximum point is (-4, -2), with the value of the function being -2 at this point.
In terms of the range, since the parabola opens upwards (because a > 0), the range extends from the y-coordinate of the minimum point (-2) to positive infinity. The range can be expressed as [-2, +∞).
The minimum value of the function is -2.