GENERAL to Transformational CHANGING FORMS GENERAL to Transformational

To General Form – Complete the squar Remember: the general form of a quadratic function is y = ax2 + bx +c The transformational form is The standard form is y = a(x-h)2 + k To GET general form: FOIL and solve for y. But how do we go FROM general form?

To General Form – Complete the squar Let’s look at an example: What is the vertex of the quadratic function y = 2x2 + 12x -4? We know that in transformational form the coefficient of ‘x’ is 1 So STEP 1 is to divide every term by ‘a’ STEP 2 is to move the non-x term over.

x2 Completing the Squar x x x x x x The right-hand side of the equation needs to be a perfect square: (x –h)2 So far we have: x2 +6x This looks like: x2 x x x x x x

x2 x2 Oops… Completing the Squar x x x x x x x To make this into a perfect square we could… x2 x Oops…

x2 x2 Completing the Squar x x x x x x x x x x x x To make this into a perfect square we could… Almost…we need to complete the square! x2 x x x x x x We were missing nine 1x1 squares.

x2 Completing the Squar x So the PERFECT SQUARE is … BUT…how did we get this? ---------- X + 3 ----------- x2 x ---------- X + 3 ----------- Added it to the right hand side. This means we must add it to both sides! We took HALF the coefficient of x and squared it. STEP 3: take half the coefficient of x, square it and add it to both sides (x+3)2

x2 Ta Da! Completing the Square x So the PERFECT SQUARE is … BUT…how did we get this? ---------- X + 3 ----------- x2 x Step 4: Factor out the 1/a term on the left hand side ---------- X + 3 ----------- We need the left hand side to have brackets. Ta Da! (x+3)2

Completing the square We started with the quadratic in general form y = 2x2 + 12x -4. The same function is in transformational form. So what is the range of y = 2x2 + 12x -4 ? Since the equivalent equation shows us a vertex of (-3,-22) with no reflection, the range is

Practice going FROM general form 1. What is the vertex of the function y = 2x2 - 8x +2? 2. What is the range of the function y = -x2 -5x +1? 3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

Practice going FROM general form 1. What is the vertex of the function y = 2x2 - 8x +2? Vertex: (2,-6)

Practice going FROM general form 2. What is the range of the function y = -x2 -5x +1? Y-value of the vertex Since there’s a reflection

Practice going FROM general form 3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

A shortcut…eventually If only there was a way to avoid having to complete the square every time to get the vertex!! Maybe there is… Let’s put the following equations into transformational form. y = 3x2 + 12x - 6 y = ax2 + bx + c STEP 1 is to divide every term by ‘a’ STEP 2 is to move the non-x term over.

A shortcut…eventually STEP 3: take half the coefficient of x, square it and add it to both sides Step 4: Factor out the 1/a term on the left hand side Who cares?

X = -b 2a Here’s the shortcut We now see that ANY general equation can be written as We haven’t finished this yet but we already see that the x-value of the vertex is X = -b 2a This is VERY important. For example: What is the axis of symmetry of the function y = 2x2 – 16x +3? To complete the square on this takes time. But

Finding y Example 2: What is the range of the graph of the function y = -x2 – 4x+ 7? We know that the x coordinate of the vertex is But how do I find y? You glade the x- value! You “plug in it, plug it in” The range is NOTE: The vertex of ANY quadratic occurs at x = -b/2a and the max or min value is: f(-b/2a)

Changing forms (quickly) Ex. Graph the parabola defined by the function: f(x) = 3x2 + 12x - 9 If f(x) = 3x2 + 12x – 9, then the vertex is We know that transformational form is best for graphing. Is it still necessary? f(-2) = 3(-2)2 +12(-2)-9 f(-2) = 3*4-24-9 f(-2) = -21 NO! Vertex (-2,-21)

Changing forms (quickly) So f(x) = 3x2 + 12x -9 can now be written: If only we knew the ‘a’ value… Since the VS = 3 we can graph from the vertex: Over 1 up 3 Over 2 up 12 Over 3 up 27

Changing Forms (quickly) What we know: f(x) = 3x2 + 12x – 9 Axis of symmetry Range (and Domain) y-intercept Max/min value But we don’t know the x-intercepts!

Finding our Roots When dealing with quadratics, we will be asked to solve quadratic equations, find the roots or find the zeros of quadratic equations or find the x-intercepts of the parabola. These all mean: solve for x. Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 But we can change this into transformational form. Since there are 2 x terms (an x2 and an x) we cannot undo what’s being done. So, in this form (for now) we’re stuck! f(x) = x2 - 2x -15 y = x2 - 2x -15 y+15 = x2 - 2x y +15 +1 = x2 - 2x +1 y +16 = (x-1)2

Finding our Roots f(x) = x2 - 2x -15 y +16 = (x-1)2 We know that for any x-intercepts: f(x) =0 y +16 = (x-1)2 0+16 = (x-1)2 To undo what’s being done to ‘x’ we need to take the square root of both sides. 16 = (x-1)2 So the 2 x-intercepts are (5,0) and (-3,0) Remember: (-4)(-4) =16 Now, this is actually 2 equations in one

Who cares? We Do Finding our Roots We’ve found the roots of a quadratic function but it involved us going back to transformational form. Is there a shortcut? When we completed the square on the general form we ended up with this Who cares? We Do If we set y= 0 we can come up with a formula to find the roots of a quadratic function. We’ll call it the “Quadratic Root Formula”

Finding our Roots Now, on the left-hand side we need to add fractions, so we need a common denominator.

TA DA!!!!! Finding our Roots So if we go back to our original example: Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 First set one side equal to 0 (since y = 0 for x-intercepts) 0 = x2 - 2x -15 Now plug in a, b and c into the quadratic root formula TA DA!!!!! So the 2 x-intercepts are (5,0) and (-3,0)

Yet another method of finding roots We have solved quadratic equations by putting it in transformational form and undoing what’s being done Making one side 0 and using the quadratic root formula Sometimes we can also factor. This uses the zero property. If (r)(s) = 0 then EITHER ‘r’ must equal 0 or ‘s’ must equal 0. This only works when the product is 0. So if we can get the quadratic function in the form (x-r)(x-s)=0 then we know that either x-r = 0 or x –s has to be zero. This is called FACTORING

Factoring to find the roots Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 Answer when the numbers are added Again, f(x) = 0 0 = x2 -2x -15 We need two numbers whose product is -15 and whose sum is -2 -5 3 ___ x ___ = -15 ___ + ___ = -2 Answer when the numbers are multiplied -5 3 So x2 -2x-15=0 (x-5)(x+3)=0 So the 2 x-intercepts are (5,0) and (-3,0) Either (x-5) = 0 Or (x+3) = 0 x = 5 x = -3

Factoring When factoring a quadratic equation where ‘a’ = 1, find two numbers that multiply to give ‘c’ and add to give ‘b’. By the way: 24 = x2 +10x 24 +25 = x2 +10x +25 49 = (x+5)2 +/- 7 = x + 5 x = -5 +7 or x = -5 -7 x = 2 or x = -12 Ex. Solve for x by factoring ___ x ___ = -24 ___ + ____ = 10 12 -2 12 -2 or x =(-10 +14)/2 =2 x = (-10-14)/2 = -12

Practice 1. Solve the following equations by factoring. a) 0= x2 + 2x +1 c) 0= x2 + 2x -24 b) 0= x2 + 5x +4 d) 0= x2 -25 2. Find the x- and y- intercepts of the following quadratics. a) f(x) = 2x2 +3x + 1 c) f(x) = x2 -5x -14 d) y= 2(x-4)2 - 32 b) f(x) = 3x2 -7x + 2 Answers 1a. x = -1 b. x = -4 or -1 c. x = -6 or 4 d. x = -5 or +5 2a. (-1,0) (-0.5, 0) (0,1) b. (1/3,0) (2,0) (0, 2) c. (-2,0) (7,0) (0, -14) d. (0,0) (8,0)