(0.5x^+6x)-3
y-intercept:
vertex is -b/2a how can i find it in this situation
2007-06-08 10:30:20 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Assume you mean: y = 0.5x^2+6x-3
The y intercept occurs when x = 0
So y- intercept = 0+0 - 3 = -3
b = 6 and a = 0.5
So -b/2a = -6/(2*.5) = -6
So the x-coordinate of the vertex is -6
So y = .5(-6)^2 +6(-6) -3
y = 18 -36 - 3 = -21
So vertex is (-6,-21)
2007-06-08 10:46:04 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
It is really unclear what the function is.
Why don't you try to write it clearly in the form:
y = some function of x
And an expression like "x^" is meaningless: Try again!
2007-06-08 10:36:11 · answer #2 · answered by ? 6 · 0⤊ 1⤋
my sister said it is 3.5
2007-06-08 10:38:56 · answer #3 · answered by *curtis* 2 · 0⤊ 1⤋
isn't it -3?
2007-06-08 10:37:44 · answer #4 · answered by Andrea 3 · 0⤊ 0⤋