SOLUTION: for the provided quadratic function of {{{y=x^2+4x+3}}} A. Find the vertex B. find the y intercept C. find the x-intercepts D. graph function

Question 331602: for the provided quadratic function of

A. Find the vertex
B. find the y intercept
C. find the x-intercepts
D. graph function
Found 2 solutions by texttutoring, solver91311:
Answer by texttutoring(324) (Show Source): You can put this solution on YOUR website!
This equation is a parabola in general form. The easiest way to graph it is to convert it to standard form, y=a(x-p)^2 + q. You can do this by completing the square:

Halve the b term (b=4 in your equation) and then square it: (4/2)^2 = 2^2 = 4

Add this number to the first 2 terms, but then subtract it at the end (so that you are just adding zero to the overall equation):

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

Now the equation is in standard form, y=a(x-p)^2 + q. The vertex is at (p,q)=(-2,-1).

You can find the y=intercept by setting x=0 and solving for y. It's easiest to do this with the general form:

y=x^2+4x+3
y = 0+)+3
y=3
The y-intercept is y=3.

To find the x-intercepts, factor, set y=0 and solve for x. You have to solve by factoring:

y=x^2+4x+3
0 = x^2+4x+3
0 = (x+3)(x+1)

Which means that the x-intercepts are x= -3, and x= -1.

You now know the x-intercepts, y-intercept, and vertex. You should be able to just connect the dots to graph the equation.

It should look like this: http://www.wolframalpha.com/input/?i=y%3Dx^2%2B4x%2B3
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

For any quadratic function with real coefficients of the form:

The vertex is located at:

Where:

The -intercept is located at

-intercepts, if they exist, are located at:

and

If the calculation under the radical, namely , is positive, then there are two -intercepts. If , then the -axis will be tangent at the vertex, i.e. the single -intercept will be (see vertex discussion above). If , then the graph does not intersect the -axis anywhere.

Because of symmetry, you can plot one additional point. At the horizontal distance from the -axis to the vertex on the OTHER side of the vertex, there will be a function value equal to the -coordinate of the -intercept. That is to say, is a point on the graph.

For your problem:

, , and

Graph: Plot the five points discussed above and draw a smooth parabolic curve through them.

John

My calculator said it, I believe it, that settles it

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