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1 § 8.4 Quadratic Functions and their Graphs 2 Graphs of Quadratic Functions The graph of a quadratic function has the shape of a parabola. But, what exactly is parabola? 3 Graphs of Quadratic Functions The graph of a quadratic function has the shape of a parabola. But, what exactly is parabola? The curve formed by the set of points in a plane that are equally distant from both a given line(called the directrix) and a given point (called the focus) that is not on the line. 4 Graphing Quadratic Functions π π₯ = π₯2 5 Graphing Quadratic Functions π π₯ = π₯2 We need to choose values for π₯ to give us π¦ values that will help us plot points on a coordinate graph. In general, choosing values around 0 helps. So numbers such as -3,-2,1,0,1,2,3. 6 Quadratics of the Form ππ₯2. 1.) If the coefficient of the term with π₯ 2 is a positive number, then the parabola will open upwards. 2.) If the coefficient of the term with π₯ 2 is a negative number, then the parabola will open downwards. This will be true always no matter what the coefficients are for the other less degree terms. 7 Quadratics of the Form ππ₯2. 1.) If the coefficient of the term with π₯ 2 is a positive number, then the parabola will open upwards. 2.) If the coefficient of the term with π₯ 2 is a negative number, then the parabola will open downwards. This will be true always no matter what the coefficients are for the other less degree terms. E.g.: Looking at π π₯ = β2π₯ 2 . 8 Graphing Quadratic Functions π π₯ = π₯2 We need to choose values for π₯ to give us π¦ values that will help us plot points on a coordinate graph. In general, choosing values around 0 helps. So numbers such as -3,-2,-1,0,1,2,3. Now looking at π π₯ = π₯ 2 β 2. 9 Graphing Quadratic Functions π π₯ = π₯2 We need to choose values for π₯ to give us π¦ values that will help us plot points on a coordinate graph. In general, choosing values around 0 helps. So numbers such as -3,-2,-1,0,1,2,3. Now looking at π π₯ = π₯ 2 β 2. Now looking at β π₯ = (π₯ β 2)2 . What is the difference? Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 10 Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 The graph of π π₯ is just the graph of π₯ 2 , but shifted to the left 3 units and raised 4 units up. 11 Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 The graph of π π₯ is just the graph of π₯ 2 , but shifted to the left 3 units and raised 4 units up. I like to have the notion of βoutsideβ and βinsideβ of the parenthesis. The 3 is βinsideβ and the 4 is βoutsideβ. 12 Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 The number that you add/subtract in the βinsideβ will shift the graph of π₯ 2 to the left/right that many units while the number that you add/subtract in the βoutsideβ will shift the graph of π₯ 2 up/down that many units. 13 14 Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 The quadratic equation here is in what is called Standard Form. Basically, it is in the form, π π₯ = π(π₯ β β)2 +π Where our π = 1, β = β3, π = 4. 15 Combinations of Transformations Letβs look at, π π₯ = (π₯ + 3)2 + 4 The quadratic equation here is in what is called Standard Form. Basically, it is in the form, π π₯ = π(π₯ β β)2 +π Where our π = 1, β = β3, π = 4. The vertex of a parabola is the point with coordinates β, π . It is basically the lowest/highest point on the parabola. 16 Axis of Symmetry The Axis of symmetry has the equation, π₯=β Where β comes from the equation π π₯ = π(π₯ β β)2 +π. It is the vertical line that basically βcuts the parabola is halfβ. 17 Vertex Formula To get the π₯-coordinate of the vertex point from the quadratic in the form ππ₯ 2 + ππ₯ + π, itβs just βπ π₯= 2π 18 Vertex Formula To get the π₯-coordinate of the vertex point from the quadratic in the form ππ₯ 2 + ππ₯ + π, itβs just βπ π₯= 2π To get your π¦-coordinate, just plug in your result for π₯ back into the quadratic equation. 19 Vertex Formula To get the π₯-coordinate of the vertex point from the quadratic in the form ππ₯ 2 + ππ₯ + π, itβs just βπ π₯= 2π To get your π¦-coordinate, just plug in your result for π₯ back into the quadratic equation. Letβs graph, π π₯ = π₯ 2 + 6π₯ + 5. 20 Number of Real Solutions ο§ If the graph of a parabola intersects the x- axis twice, then the quadratic equation has 2 real solutions. ο§ If the graph of a parabola intersects the xaxis at a single point, then the quadratic equation has 1 real solution. ο§ If the graph of a parabola does not intersect the x-axis at all, then the quadratic equation has no real solutions. 22 Graphing functions of the form ππ₯2 + ππ₯ + π by Completing the Square We will need to use completing the square for a quadratic function of the form ππ₯ 2 + ππ₯ + π to get it in the form π(π₯ β β)2 +π. E.g.: 3π₯ 2 β 12π₯ + 8 23 Sketch the graph π π₯ = β2π₯ 2 + 12π₯ β 10 1.) Find the coordinates of the vertex. 2.) What is the axis of symmetry? 3.) Is it going to open upward or downward? 4.) Choose values for x around the h value of the vertex and make a βT tableβ to get points. 24 Sketch the graph π π₯ = β2π₯ 2 + 12π₯ β 10 1.) Find the coordinates of the vertex. 2.) What is the axis of symmetry? 3.) Is it going to open upward or downward? 4.) Choose values for x around the h value of the vertex and make a βT tableβ to get points. Also, recall, that to find the x-intercepts, set the quadratic function equal to 0 and to find the yintercept, set the xβs equal to 0.