Fireworks – From Standard to Vertex Form • What's the pattern? + x 6 x2 6x 36 (x + 6)2 x2 + 12x + 36 • How about these? x2 + 4x ______ (x _____ )2 + 4 + 2 x2 + 10x ______ (x _____ )2 + 25 + 5 x2 – 14x ______ (x _____ )2 + 49 – 7

Fireworks – From Standard to Vertex Form • Converting from standard form to vertex form can be easy… x2 + 6x + 9 (x + 3)2 x2 – 2x + 1 = (x – 1)2 x2 + 8x + 16 = (x + 4)2 x2 + 20x + 100 = (x + 10)2 … but we're not always so lucky

Fireworks – From Standard to Vertex Form • The following equation requires a bit of work to get it into vertex form. y = x2 + 8x + 10 y = (x2 + 8x ) + 10 + 16 – 16 16 is added to complete the square. 16 is sub-tracted to maintain the balance of the equation. y = (x + 4)2 – 6 The vertex of this parabola is located at ( –4, –6 ).

Fireworks – From Standard to Vertex Form • Lets do another. This time the x2 term is negative. y = –x2 + 12x – 5 Un-distribute a negative so that when can complete the square y = (–x2 + 12x ) – 5 y = –(x2 – 12x ) – 5 y = –(x2 – 12x ) – 5 + 36 + 36 The 36 in parentheses becomes negative so we must add 36 to keep the equation balanced. y = – (x – 6)2 + 31 The vertex of this parabola is located at ( 6, 31 ).

Fireworks – From Standard to Vertex Form • The vertex is important, but it's not the only important point on a parabola y-intercept at (0, 10) x-intercepts at (1,0) and (5, 0) Vertex at (3, -8)

Fireworks – From Standard to Vertex Form • In addition to telling us where the vertex is located the vertex form can also help us find the x-intercepts of the parabola. Just set y = 0, and solve for x. y = (x + 4)2 – 6 0 = (x + 4)2 – 6 Add 6 to both sides 6 = (x + 4)2 Take square root of both sides = x + 4 = x + 4 Subtract 4 from both sides –1.551 = x –6.449 = x x-intercepts at –1.551 and -6.449

Fireworks – From Standard to Vertex Form • Another example, this time the parabola is concave down. y = –(x – 7)2 + 3 0 = –(x – 7)2 + 3 Subtract 3 from both sides –3 = –(x – 7)2 Divide both sides by -1 3 = (x – 7)2 Take square root of both sides 1.732 = x – 7 –1.732 = x – 7 Add 7 to both sides 8.732 = x 5.268 = x x-intercepts at 5.268 and 8.732

Fireworks – From Standard to Vertex Form • Another example, this time the a value is 0.5. y = 0.5(x + 3)2 + 5 0 = 0.5(x + 3)2 + 5 Subtract 5 from both sides –5 = 0.5(x + 3)2 Divide both sides by 0.5 –10 = (x + 3)2 Take square root of both sides = x + 3 = x + 3 Subtract 3 from both sides Error = x Error = x NO x-intercepts… can't take square root of a negative number.

Fireworks – From Standard to Vertex Form • Find the x-intercepts of the parabola for each of the quadratic equations. 1. y = (x – 7)2 – 9 x-intercepts at 10 and 4 2. y = 3(x + 4)2 + 6 NO x-intercepts 3. y = –0.5(x – 2)2 + 10 x-intercepts at 6.472 and –2.472 • Is there a way to tell how many x-intercepts a parabola will have without doing any calculations?

Fireworks – From Standard to Vertex Form • Finding the y-intercept is a little more straightforward. Just set x = 0 and solve for y. y = (x + 4)2 – 6 y = (0 + 4)2 – 6 y-intercept at (0, 10) y = 10 • The quadratic equation does not have to be vertex form to find the y-intercept. y = x2 + 8x + 10 y-intercept at (0, 10) y = (0)2 + 8(0) + 10 y = 10