Help w/ quadratic equation

[Spercy55]

Hello, Can someone pleaseeeee help me understand how to do this problem. It says given the following equation y= X squared +2x- 6 ffind the following:

a. vertex
b. x-intercepts
c. y intercept
d. sketch of graph


How do I start?

Thanks
Cindy

 

[Loren]

Hello, Can someone pleaseeeee help me understand how to do this problem. It says given the following equation y= X squared +2x- 6 ffind the following:

a. vertex <<< x[sup:16vdiv8y]2[/sup:16vdiv8y] = y is a parabola with vertex at (0,0). (x-h)[sup:16vdiv8y]2[/sup:16vdiv8y] = (y-k) is the same parabola but having been shifted h units horizontally and k units vertically. So, you need to rewrite your equation, getting it into that form. Get the x terms on the left side and complete the square. The y term and appropriate constant will be on the right side. If we had the equation y = x[sup:16vdiv8y]2[/sup:16vdiv8y] + 4x + 10, we would take the following steps.
y = x[sup:16vdiv8y]2[/sup:16vdiv8y] + 4x +10
x[sup:16vdiv8y]2[/sup:16vdiv8y] + 4x + 10 = y
x[sup:16vdiv8y]2[/sup:16vdiv8y] + 4x = y - 10
x[sup:16vdiv8y]2[/sup:16vdiv8y] + 4x + 4 = y - 10 + 4
(x+2)[sup:16vdiv8y]2[/sup:16vdiv8y] = y - 6 <<<This tells you that the vertex is at (-2,6)

b. x-intercepts <<< This is where the curve crosses the x-axis. Set the value of y to zero and solve for x. Those are the points on the x-axis where the curve crosses it.

c. y intercept <<< This is where the curve crosses the y-axis. Set the value of x to zero and solve for y. That's the point at which it crosses the y-axis.

 

[fasteddie65]

As far as graphing, once you know the vertex is (-2, 6), choose x-values on either side of -2: -4, -3, -1, and 0.
Find the corresponding y-values for each of these x-values. That will give you 5 pairs of points to graph, sufficient to make a very accurate graph.

 

[mmm4444bot]

For those who are interested, there is another method to find the coordinates of the vertex point.

We can use a formula to determine the x-coordinate of this point; after we find the x-coordinate, we get the corresponding y-coordinate by substitution. I'll show an example.

The formula for the x-coordinate of the vertex point is easy to remember because it's the same as the Quadratic Formula with the radical term (i.e., the plus or minus square root part) removed.

Given a quadratic equation in the form: y = Ax^2 + Bx + C

x-coordinate of the vertex point = -B/(2A)

EG: y = 14x^2 - 10x + 33

A = 14 and B = -10

x-coordinate of the vertex point = 10/(2*14)

= 10/28

= 5/14

We now find the corresponding y-coordinate by substitution into the given equation.

y = 14x^2 - 10x + 33

y = 14(5/14)^2 - 10(5/14) + 33

= (25 - 50 + 462)/14

= 437/14

The coordinates of the vertex point are (5/14, 437/14).