What value of x will give the minimum value for 9x^2 + 18x + 7?
y = 9x^2 + 18x + 7
IF you do calculus:
slope = dy /dx = 18 x + 18
where is it zero?
18 x = - 18
x = -1
IF you do algebra:
y = 9x^2 + 18x + 7
9 x^2 + 18 x = y-7
x^2 + 2x = (y-7)/9
x^2 + 2 x + 1 = (y-7)/9 + 9/9
(x+1)^2 = (1/9)(y +2/9)
vertex at x = -1
typo
x+1)^2 = (1/9)(y + 2)
vertex at x = -1
Y = 9x^2+18x+7
h = Xv = -B/2A = -18/18 = -1.
So X = -1 gives minimum value of the Eq. :
Y min. = 9*(-1)^2+18*(-1)+7 = -2 = min. value of Eq.
To find the minimum value of the quadratic function 9x^2 + 18x + 7, we need to determine the x-value that corresponds to the vertex of the parabola.
The vertex of a quadratic function in the form ax^2 + bx + c can be found using the formula x = -b / (2a).
In this case, a = 9, b = 18, and c = 7.
So, we substitute these values into the formula:
x = -18 / (2 * 9)
x = -18 / 18
x = -1
Therefore, the value of x that will give the minimum value for the quadratic function 9x^2 + 18x + 7 is x = -1.