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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
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-> 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
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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)
(
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):
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(20879)
(
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):
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