• Aug 26th 2007, 04:31 PM
casper
I really need some help here. Please be specific as this is new to me.

Graph the quadratic equation after completing the given table of values.
y = x^2 + 2x – 3.
Include the axis of symmetry and show specific points, vertex, and intercepts.
x, y
–3, ?
-2, ?
–1, ?
0, ?
1, ?

I don't expect anyone to try to tell me how to graph the points. I can do that. I need step by step with how to get the points, axis of symmetry and vertex. My text doesn't really help me understand and this is the only problem I can't figure out of my assignment.
Greatly appreciated!! :(:confused::(:confused:
• Aug 26th 2007, 04:40 PM
Jhevon
Quote:

Originally Posted by casper
I really need some help here. Please be specific as this is new to me.

Graph the quadratic equation after completing the given table of values.
y = x^2 + 2x – 3.
Include the axis of symmetry and show specific points, vertex, and intercepts.
x, y
–3, ?
-2, ?
–1, ?
0, ?
1, ?

I don't expect anyone to try to tell me how to graph the points. I can do that. I need step by step with how to get the points, axis of symmetry and vertex. My text doesn't really help me understand and this is the only problem I can't figure out of my assignment.
Greatly appreciated!! :(:confused::(:confused:

you get the points by plugging in the given x-values into the formula given

for instance, when x = -3, y = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0

this is an upward-opening parabola, the axis of symmetry is the vertical line passing through the vertex

the x-value for the vertex is given by: $x = \frac {-b}{2a}$, where b is the coefficient of x and a is the coefficient of x^2

to find the y-coordinate for the vertex, you again would plug in the x-value into the formula
• Aug 26th 2007, 05:33 PM
Krizalid
Quote:

Originally Posted by casper
y = x^2 + 2x – 3.

To get quick the vertex

$y=x^2+2x-3=(x+1)^2-4$

So when $x=-1,\,y=-4\,\therefore\,\mathcal V=(-1,-4)\,\blacksquare$