Hi

Note: vertex form of a parabola,  where(h,k) is the vertex

y=1/2(x+3)^2-1  |a = .5 > 0  Parabola opens upward.

Vertex(-3,-1) Line of symmetry is x = -3

 y = .5*9-1 when x = 0  y-intercept is (0,7/2)

 0 = .5(x+3)^2 - 1

 2 = (x+3)^2

 ±-3 = x when y = 0  x-intercepts are (-4.4142,0) and (-1.5858, 0)









ewatrrr@aol.com

private/pay tutoring Inquiries only, please. 

 

RELATED QUESTIONS

we're doing parabolas at the moment. I need to know how to find the turning point of a... (answered by josgarithmetic)

How do you find the vertex, axis of symmetry and graph: f(x) -x^2-6x+3, in the back of my  (answered by stanbon,jim_thompson5910,Earlsdon)

i need help and a couple of things explained better if possible

this is what my... (answered by richwmiller)

I do not know how to find the x-intercept and the vertex of a quadratic equation. For... (answered by scott8148)

I am lost on what to do with is the given data. The question is;
Determine the vertex of  (answered by stanbon)

The function f(t) = t2 + 6t − 20 represents a parabola.

Part A: Rewrite the... (answered by MathLover1)

How do you solve for x when it acts as the exponent?
Example:... (answered by josgarithmetic)

How do you compose an equation from a quadratic graph?  The vertex is (-1,-9) The... (answered by scott8148)

y=x^2+4x-5 

find the x-intercept

find the y intercept

find the vertex 

find... (answered by josgarithmetic)