Graphing Quadratic Functions Chapter 5.1 Graphing Quadratic Functions

Quadratic Function A function of the form y=ax2+bx+c where a≠0 making a u-shaped graph called a parabola. Example quadratic equation:

Vertex- Axis of symmetry- The lowest or highest point of a parabola. The vertical line through the vertex of the parabola. Axis of Symmetry

Find the Vertex - = a b 1 4 , x b a = - 2 ( - x = 4 2 1 )( ) x = 2 y x To find the x coordinate of the vertex, use the equation Then substitute the value of x back into the equation of the parabola and solve for y. You are given the equation y=-x2 + 4x –1. Find the coordinates of the vertex. - = a b 1 4 , 2 y x = - + 4 1 x b a = - 2 y = - + – 2 4 1 ( ) y = - + 4 8 1 ( - x = 4 2 1 )( ) y = 3 The coordinates of the vertex are (2,3) Substitute and solve for y x = 2

Table of Values Choose two values of x that are to the right or left of the x-coordinate of the vertex. Substitute those values in the equation and solve for y. Graph the points. (Keep in mind the value of a as this will help you determine which way the graph opens.) Since a parabola is symmetric about the vertical line through the vertex, you can plot mirror image points with the same y-values on the “other side” of the parabola. x y = -x2 + 4x – 1 y y = -(1)2 + 4(1) – 1 y = -1 +4 – 1 1 2 y = -(-1)2 + 4(-1) – 1 y = -1 – 4 – 1 -6 -1

Graph the Parabola x y Plot the vertex and the points from your table of values: (2,3), (1,2), (-1,-6). Use the symmetry of parabolas to plot two more points on the “other side” of the graph. The point (1,2) is one unit away from the line of symmetry, so we can also plot the point (3,2). The point (-1,-6) is three units away from the line of symmetry, so we can also plot the point (5,-6). Sketch in the parabola.

You Try It Find the vertex of the following quadratic equations. Make a table of values and graph the parabola.

Problem 1 y x The vertex is at (2,-4) Notice, a is positive, so the graph opens up. y x The vertex is at (2,-4)

Problem 2 y x The vertex is at (0,3) Notice, a is negative, so the graph opens down. x y The vertex is at (0,3)

Problem 3 y x The vertex is at (3,-5) Notice, a is positive, so the graph opens up. x y The vertex is at (3,-5)

Vertex Form Equation y=a(x-h)2+k If a is positive, parabola opens up If a is negative, parabola opens down. The vertex is the point (h,k). The axis of symmetry is the vertical line x=h. Don’t forget about 2 points on either side of the vertex! (5 points total!)

Vertex Form (x – h)2 + k – vertex form Each function we just looked at can be written in the form (x – h)2 + k, where (h , k) is the vertex of the parabola, and x = h is its axis of symmetry. (x – h)2 + k – vertex form Equation Vertex Axis of Symmetry y = x2 or y = (x – 0)2 + 0 (0 , 0) x = 0 y = x2 + 2 or y = (x – 0)2 + 2 (0 , 2) y = (x – 3)2 or y = (x – 3)2 + 0 (3 , 0) x = 3

Example 1: Graph y = (x + 2)2 + 1 Analyze y = (x + 2)2 + 1. Step 1 Plot the vertex (-2 , 1) Step 2 Draw the axis of symmetry, x = -2. Step 3 Find and plot two points on one side , such as (-1, 2) and (0 , 5). Step 4 Use symmetry to complete the graph, or find two points on the left side of the vertex.

Your Turn! Analyze and Graph: y = (x + 4)2 - 3. (-4,-3)

Example 2: Graph y= -.5(x+3)2+4 a is negative (a = -.5), so parabola opens down. Vertex is (h,k) or (-3,4) Axis of symmetry is the vertical line x = -3 Table of values x y -1 2 -2 3.5 -3 4 -4 3.5 -5 2 Vertex (-3,4) (-4,3.5) (-2,3.5) (-5,2) (-1,2) x=-3

Table of values with 4 points (other than the vertex? Now you try one! y=2(x-1)2+3 Open up or down? Vertex? Axis of symmetry? Table of values with 4 points (other than the vertex?

(-1, 11) (3,11) X = 1 (0,5) (2,5) (1,3)

Intercept Form Equation y=a(x-p)(x-q) The x-intercepts are the points (p,0) and (q,0). The axis of symmetry is the vertical line x= The x-coordinate of the vertex is ------- To find the y-coordinate of the vertex, plug the x-coord. into the equation and solve for y. If a is positive, parabola opens up If a is negative, parabola opens down.

Example 3: Graph y=-(x+2)(x-4) Since a is negative, parabola opens down. The x-intercepts are (-2,0) and (4,0) To find the x-coord. of the vertex, use To find the y-coord., plug 1 in for x. Vertex (1,9) The axis of symmetry is the vertical line x=1 (from the x-coord. of the vertex) (1,9) (-2,0) (4,0) x=1

Now you try one! y=2(x-3)(x+1) Open up or down? X-intercepts? Vertex? Axis of symmetry?

x=1 (-1,0) (3,0) (1,-8)

Changing from vertex or intercepts form to standard form The key is to FOIL! (first, outside, inside, last) Ex: y=-(x+4)(x-9) Ex: y=3(x-1)2+8 =-(x2-9x+4x-36) =3(x-1)(x-1)+8 =-(x2-5x-36) =3(x2-x-x+1)+8 y=-x2+5x+36 =3(x2-2x+1)+8 =3x2-6x+3+8 y=3x2-6x+11

Challenge Problem Write the equation of the graph in vertex form.