Quadratic Functions

[Guest]

I'm working with my son on Grahping a Quaratic Funtion and it's been going great until I check this one question...
y=-10x(Squared) + 12 x

We are using
x = -2 got y (-64)
x= -1 got y( -22)
x= 0 got y(0)
x = 1 got y(7)
x-=2 got y(-16)

So I thought the vertex would be (0,0) and x=0
in the back of the book it had(3/5, 3 and 3/5) x= 3/5?????
What am I doing wrong???

 

[soroban]

Hello, 4 little piggies mom!

\(\displaystyle y\:=\:-10x^2\,+\,12x\)

We are using:
\(\displaystyle x\,=\,\)-\(\displaystyle 2\;\;\Rightarrow\;\;y\,=\,\)-\(\displaystyle 64\)
\(\displaystyle x\,=\,\)-\(\displaystyle 1\;\;\Rightarrow\;\;y\,=\,\)-\(\displaystyle 22\)
\(\displaystyle x\,=\,0\;\;\Rightarrow\;\;y\,=\,0\)
\(\displaystyle x\,=\,1\;\;\Rightarrow\;\;y\,=\,\)2 \(\displaystyle \;\) . . . typo?
\(\displaystyle x\,=\,2\;\;\Rightarrow\;\;y\,=\,\)-\(\displaystyle 16\)

So I thought the vertex would be (0,0) and \(\displaystyle x\,=\,0\;\) . . . no

In the back of the book it had \(\displaystyle \left(\frac{3}{5},\,3\frac{3}{5}\right)\)

What am I doing wrong?

You seem to think that the vertex is always at \(\displaystyle x\,=\,0\) . . . why?


I don't suppose you plotted those points, did you?

                    |
                    |     *
                    |   (1,2)
      - + - - + - - * - - + - - + - -
                  (0,0)
                    |           *
              *     |         (2,-16)
          (-1,-22)  |
                    |
                    |
        *           |
    (-2,-64)        |

It certainly doesn't look like (0,0) is the vertex, does it?


You're expected to know the "Vertex Formula".

For the quadratic function: \(\displaystyle \,y\;=\;ax^2\,+\,bx\,+\,c\)
\(\displaystyle \;\;\;\)the vertex is found at: \(\displaystyle \,x\:=\;\frac{-b}{2a}\)

Your function has: \(\displaystyle \,a\,=\,10,\;b\,=\,12\)
\(\displaystyle \;\;\)your vertex is at: \(\displaystyle \,x\:=\;\frac{-12}{2(-10)}\:=\:\frac{12}{20}\:=\:\frac{3}{5}\)

Then: \(\displaystyle \,y\:=\:-10\left(\frac{3}{5}\right)^2\,+\,12\left(\frac{3}{5}\right)\:=\:\frac{18}{5}\)

Got it?