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:
\ =\ ax^2\ +\ bx +\ c)
The vertex is located at:
)
Where:
\ =\ p\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right) +\ c)
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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