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
You have the correct min.
Isn't this the same question as this one?
w-ww-.-jiskha.-c-om/display-.cgi?id=-1207258851
Err exactly the same. The min is wrong >.<. The minimum value occurs at x = -2 but the minimum value is not -2.
Your parabola in standard form is
y = 2(x+2)^2 - 10 , (check it by expanding this)
so you have a parabola opening upwards with vertex at (-2,-10)
Minimum value is -10, when x=-2
axis of symmetry is x=-2
domain: any real number for x
range: y ≥ -10, y any real number
To find the minimum or maximum point of a quadratic function, you need to calculate the coordinates of the vertex. The vertex is the highest or lowest point on the parabola.
For the given function f(x) = 2x^2 + 8x - 2, we can find the vertex using the formula x = (-b)/(2a), where a, b, and c are the coefficients of the quadratic equation (in the form ax^2 + bx + c).
In this case, a = 2 and b = 8. Plugging these values into the formula, we get:
x = (-8)/(2 * 2) = -8/4 = -2
So the x-coordinate of the vertex is -2. To find the y-coordinate, we substitute this value back into the original equation:
f(-2) = 2(-2)^2 + 8(-2) - 2
f(-2) = 2(4) - 16 - 2
f(-2) = 8 - 16 - 2
f(-2) = -10
Therefore, the y-coordinate of the vertex is -10. The vertex is (-2, -10).
The axis of symmetry is a line that divides the parabola into two equal halves. It passes through the vertex and is vertical when the parabola opens horizontally, like in this case. The equation of the axis of symmetry is given by x = -2.
The range is the set of all possible y-values of the function. In this case, the parabola opens upwards, so the minimum value of the function is at the vertex, which is -10. The range is [-10, infinity).
Therefore, the minimum point is (-2, -10) and the minimum value is -10.