SOLUTION: Someone PLEASE help me! I am solving and graphing parabolas and I just dont understand the steps involved....I also need to graph each one and find the vertex's and intercepts?? I

Question 111879: Someone PLEASE help me! I am solving and graphing parabolas and I just dont understand the steps involved....I also need to graph each one and find the vertex's and intercepts?? I dont know what to do first...no has explained this to me so I can understand. Here is an example of what I am working on: f(x)=-3x^2 + x -5
I would appreciate any help!
Thanks so much!
Bonnie Cook
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First let's find the vertex. To find the vertex, we first need the axis of symmetry (ie the x-coordinate of the vertex)

To find the axis of symmetry, use this formula:

From the equation we can see that a=-3 and b=1

Plug in b=1 and a=-3

Multiply 2 and -3 to get -6

Reduce

So the axis of symmetry is

So the x-coordinate of the vertex is . Lets plug this into the equation to find the y-coordinate of the vertex.

Lets evaluate

Start with the given polynomial

Plug in

Square to get

Multiply

Reduce

Combine like terms

So the y-coordinate of the vertex is

So the vertex is

Now let's find the x-intercepts. To find the x-intercepts, let y=0 and solve for x

Set y equal to zero to solve for x

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=-3, b=1, and c=-5

Square 1 to get 1

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and -3 to get -6

After simplifying, the quadratic has roots of

or

Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

Start with the given equation

Add to both sides

Factor out the leading coefficient

Take half of the x coefficient to get (ie ).

Now square to get (ie )

Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation

Now factor to get

Distribute

Multiply

Now add to both sides to isolate y

Combine like terms

Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.

Check:

Notice if we graph the original equation we get:

Graph of . Notice how the vertex is (,).

Notice if we graph the final equation we get:

Graph of . Notice how the vertex is also (,).

So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.

as you can see on graph above, there are no
now, use the quadratic formula to solve for to show that this function has

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=-3, b=1, and c=-5

Square 1 to get 1

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and -3 to get -6

After simplifying, the quadratic has roots of

or

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