Question

Where is the vertex of the parabola?
y=x^2x+3

Answer 1

vertex of y=ax^2+bx+c is given by
(-b/(2a), f(-b/(2a))

Answer 2

so x = -b/(2a) then plug that back into f(x) to find y..

Answer 3

no calc needed....

Answer 4

So it's above the X-axis

Answer 5

x = -b/2a = -1/2(1) = -1/2, and so y = (-1/2)^2+(-1/2) + 3 = 11/4
so
(-1/2, 11/4)

Answer 6

the question should say x^2+x+3?

Answer 7

\(\bf y=x^2x+3\ ?\qquad y=x^2+3\ ?\)

Answer 8

im guessing there is a missing + or -

Answer 9

@Abbii_15 what should the question say?

Answer 10

if it should say
\(x^2+x+3\)

then
\(x = -b/2a = -1/2(1) = -1/2\) and so \(\\y = (-1/2)^2+(-1/2) + 3 = 11/4\)
so
\(\large V = (-1/2, 11/4)\)

Answer 11

shes not in calc, I can tell by the other question.