2Graphing the parabola y = f (x) = ax2 Consider the equation y = x2Axis of symmetry: x = ( y=x2 is symmetric with respect to the y-axis )x– 2– 112y4411(–2, 4)(–1, 1)(0, 0)(1, 1)(2, 4)When a > 0, the parabola opens upwards and is called concave up.The vertex is called a minimum point.yVertex(0, 0)x
3For the function equation y = x2 , what is a ? a = 1 . What if a does not equal 1?Consider the equation y = – 4x2 .What is a ?a = – 4x– 2– 112yy– 4– 4– 16– 16(–2, –16)(–1, –4)(0, 0)(1, –4)(–4, –16)xWhen a < 0, the parabola opens downward and is called concave down.The vertex is a maximum point.
4Properties of the Parabola f (x) = ax2 The graph of f (x) = ax2 is a parabola with the vertex at the origin and the y axis as the line of symmetry.If a is positive, the parabola opens upward, if a is negative, the parabola opens downward.If a is greater than 1 (a > 1), the parabola is narrower then the parabola f (x) = x2.If a is between 0 and 1 (0 < a < 1), the parabola is wider than the parabola f (x) = x2.
5Graphing the parabola y = f (x) = ax2 + k Consider the equation y = – 4x2 – 3 . What is a ?a = – 4Graphical Approach:Numerical Approach:x– 2– 112-4x2-16-4-4x2-3-19-7-3xVertex(0, -3)Algebraic Approach: y = – 4x2 – 3
6The graph y = – 4x2 is shifted down 3 units. Vertex(0, -3)y = – 4x2 – 3 .In general the graph of y = ax2 + k is the graph of y = ax2 shifted vertically k units. If k > 0, the graph is shifted up. If k < 0, the graph is shifted down. (P. 267)
7Consider the equation y = – 4(x – 3)2 . What is a ? a = – 4. What effect does the 3 have on the function?The axis of symmetry is x = 3.Numerical Approach:yx– 2– 112–4 x2-16-43-36x-4 (x-3) 2-100-64-36-16-4Axis of symmetry is shifted 3 units to the right and becomes x = 3y= – 4x2y= – 4(x – 3)2
8Vertex Form of a Quadratic Function The quadratic functionf(x) = a(x – h)2 + k, aThe graph of f is a parabola .Axis is the vertical line x = h.Vertex is the point (h, k).If a > 0, the parabola opens upward.If a < 0, the parabola opens downward.
96.6 Analyzing Graphs of Quadratic Functions A family of graphs – a group of graphs that displays one or more similar characteristics.Parent graph – y = x2Notice that adding a constant to x2 moves the graph up.Notice that subtracting a constant from x before squaring it moves the graph to the right.y = x2 + 2y = (x – 3)2y = x2
10Vertex 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 formEquationVertexAxis of Symmetryy = x2 or y = (x – 0)2 + 0(0 , 0)x = 0y = x2 + 2 or y = (x – 0)2 + 2(0 , 2)y = (x – 3)2 or y = (x – 3)2 + 0(3 , 0)x = 3
11Graph Transformations As the values of h and k change, the graph of y = a(x – h)2 + k is the graph of y = x2 translated.| h | units left if h is negative, or |h| units right if h is positive.| k | units up if k is positive, or | k | units down if k is negative.
12Graph a Quadratic Function in Vertex Form Analyze y = (x + 2) Then draw its graph.This function can be rewritten as y = [x – (-2)]2 + 1.Then, h = -2 and k = 1The vertex is at (h , k) = (-2 , 1), the axis of symmetry is x = -2. The graph is the same shape as the graph of y = x2, but is translate 2 units left and 1 unit up.Now use the information to draw the graph.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 of the axis symmetry, such as (-1, 2) and (0 , 5).Step 4 Use symmetry to complete the graph.
13Graph Transformations How does the value of a in the general form y = a(x – h)2 + k affect a parabola? Compare the graphs of the following functions to the parent function, y = x2.a.b.c.d.aby = x2dc
14Since you know the vertex, use vertex form! y=a(x-h)2+k Example: Write a quadratic function for a parabola with a vertex of (-2,1) that passes through the point (1,-1).Since you know the vertex, use vertex form! y=a(x-h)2+kPlug the vertex in for (h,k) and the other point in for (x,y). Then, solve for a.-1=a(1-(-2))2+1-1=a(3)2+1-2=9aNow plug in a, h, & k!
15Standard 9Write a quadratic function in vertex formVertex form- Is a way of writing a quadratic equation that facilitates finding the vertex.y – k = a(x – h)2The h and the k represent the coordinates of the vertex in the form V(h, k).The “a” if it is positive it will mean that our parabola opens upward and if negative it will open downward.A small value for a will mean that our parabola is wider and vice versa.15
16( ) Standard 9 Write a quadratic function in vertex form Write y = x2 – 10x + 22 in vertex form. Then identify the vertex.y = x2 – 10x + 22Write original function.y + ? = (x2 –10x + ? ) + 22Prepare to complete the square.Add–102()=(–5)25to each side.y + 25 = (x2 – 10x + 25) + 22y + 25 = (x – 5)2 + 22Write x2 – 10x + 25 as a binomial squared.y + 3 = (x – 5)2Write in vertex form.The vertex form of the function is y + 3 = (x – 5)2. The vertex is (5, –3).ANSWER16
17Write a quadratic function in vertex form EXAMPLE 1Write a quadratic function in vertex formWrite a quadratic function for the parabola shown.SOLUTIONUse vertex form because the vertex is given.y – k = a(x – h)2Vertex formy = a(x – 1)2 – 2Substitute 1 for h and –2 for k.Use the other given point, (3, 2), to find a.2 = a(3 – 1)2 – 2Substitute 3 for x and 2 for y.2 = 4a – 2Simplify coefficient of a.1 = aSolve for a.
18EXAMPLE 1Write a quadratic function in vertex formANSWERA quadratic function for the parabola is y = (x – 1)2 – 2.
19EXAMPLE 1Graph a quadratic function in vertex form14Graph y – 5 = – (x + 2)2.SOLUTIONSTEP 1Identify the constants a = – , h = – 2, and k = 5.Because a < 0, the parabola opens down.14STEP 2Plot the vertex (h, k) = (– 2, 5) and draw the axis of symmetry x = – 2.
20EXAMPLE 1Graph a quadratic function in vertex formSTEP 3Evaluate the function for two values of x.x = 0: y = (0 + 2)2 + 5 = 414–x = 2: y = (2 + 2)2 + 5 = 114–Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.STEP 4Draw a parabola through the plotted points.
21GUIDED PRACTICEfor Examples 1 and 2Graph the function. Label the vertex and axis of symmetry.1. y = (x + 2)2 – 32. y = –(x + 1)2 + 5