Vertex form is designed to easily yield the vertex of a quadratic function.\u00a0 A quadratic function of the form<\/p>\n
y=a(x-p)\u00a0\u00b2 + q<\/span><\/p>\n has vertex at (p,q).<\/strong><\/p>\n Example 1:\u00a0 Find the vertex of y=-3(x-4)\u00b2 +9<\/span><\/p>\n Solution: the vertex is at (4,9)<\/strong>.<\/p>\n Notice the (“opposite, same”) pattern: the x-coordinate is opposite to what you see in the brackets, whereas the y-coordinate is the same as what you see added (or subtracted) at the end.<\/p>\n Example 2: Find the vertex of y=2(x+5)\u00b2 -3<\/span><\/p>\n Solution: the vertex is at (-5,-3).<\/strong><\/p>\n Notice that the number multiplying in front of the brackets does not affect the vertex.<\/p>\n Example 3: Find the vertex of y=(x-5)\u00b2<\/span><\/p>\n Solution: Remembering the form y=a(x-p)\u00b2 +q<\/span>, we need to discern the values of p and q. Clearly, p=5. q=0, because<\/p>\n y=(x-5)\u00b2<\/span><\/p>\n can also be written as<\/p>\n y=(x-5)\u00b2 + 0<\/span>.<\/p>\n Therefore, the vertex is at (5,0).<\/strong><\/p>\n Example 4: Find the vertex of y=3x\u00b2 + 7<\/span><\/p>\n Solution: Going back to y=a(x-p) \u00b2 + q<\/span>, we realize that although q=7, we seem to be missing p. However, we can rewrite our equation as y=3(x-0)\u00b2 + 7<\/span>. Now, we realize that p=0. The vertex is at (0,7).<\/strong><\/p>\n Identifying the vertex can be tricky, but I hope this helps.<\/p>\n