anonymous
  • anonymous
For the quadratic function, finds it's vertex and line of symmetry, then graph the function. 3x^2 - 18x +32.
Algebra
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katieb
  • katieb
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amistre64
  • amistre64
it vertex will be on its line of symmetry
amistre64
  • amistre64
given a quad of the form: ax^2 + bx + c the line of symmetry can be formulated as: x = -b/2a by plugging value of x into the eqaution, we can solve for y
anonymous
  • anonymous
Ok I am stuck at this \[\frac{ 18 \pm \sqrt{-60} }{ 6}\] I know it will be \[3 \pm \sqrt{-60}\] but do I change that -60 to \[2\sqrt{5i}\] over 6?

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anonymous
  • anonymous
Ok I guess what I'm asking is my vertex going to be \[3 + \frac{ 2\sqrt{5i} }{ 6 }, 3 - \frac{ 2\sqrt{5i} }{ 6 }\]
amistre64
  • amistre64
there are 2 parts to the formula you are trying to use. there is the axis of symmetry part and the distance from the axis |dw:1371229128134:dw|
amistre64
  • amistre64
notice that the vertex is on the line x= -b/2a
amistre64
  • amistre64
in this case: x = 3
amistre64
  • amistre64
to find the y value of your vertex; plug in x=3 into the equation to solve for y
amistre64
  • amistre64
thinkof it like this: the vertex of your head. the zero are what you get when you strch your arms out as far as then can go does the end of the finger tips become your head?
anonymous
  • anonymous
ok no you're not helping but thank you. This whole chapter is asking way different questions then what you're showing me nothing that what you said looks like anything I just did on my homework.
amistre64
  • amistre64
For the quadratic function, finds it's vertex and line of symmetry your attempt at using the quadratic function in its entirety is flawed. the formula for the line of symmetry is used in the quadratic: it is the \[x=-\frac b{2a}\]part of it\[x=\frac{-(-18)}{2(3)}=\frac{18}{6}=3\]
amistre64
  • amistre64
the vertex is a point that is on the line of symmetry the vertex is the point (x,y) such that x=3 y = 3(3)^2 - 18(3) +32

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