given y=(x-3)(x+2)

A) State the x-intercepts

B) State the x-coordinate of the vertex

C) Calculate the y-coordinate of the vertex

D) Sketch the relation

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- Aug 7th 2008, 08:56 AMeuclid2quadration relations
given y=(x-3)(x+2)

A) State the x-intercepts

B) State the x-coordinate of the vertex

C) Calculate the y-coordinate of the vertex

D) Sketch the relation - Aug 7th 2008, 11:40 AMChop Suey
A) The x-intercepts are the points the graph cross when y = 0. So if

$\displaystyle 0 = (x-3)(x+2)$

What are the x-intercepts now?

B and C) Okay, here you're gonna have to multiply both expressions, then complete the square. Do you know how to complete the square? In the end, you'll get an expression (after completing the square) like this:

$\displaystyle y = a(x-h)^2 + k$

The vertex is (x,y) = (h,k)

Be careful! If what you get in the end is $\displaystyle a(x+h)^2 + k$ remember that it's actually $\displaystyle a(x-(-h))^2+ k$, so the vertex (x,y) = (-h, k)

D) The relation is a parabola. You got the x-intercepts and the vertex, now plot them and sketch a parabola that contains these points.

If there is something you don't understand, don't hesitate to say so. - Aug 7th 2008, 11:51 PMbadgerigar
You can also find the x-coordinate of the vertex by using the fact that it is half-way between the x-intercepts. The y-coordinate can then be found by substituting in the x-coordinate.

Edit:typo in my typo correction :S

By the way, there is a typo in Chop-Suey's post; after completing the square the expression should be in the form $\displaystyle y = a(x-h)^2+k$