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10.2 Quadratic Functions

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**10.2 – Quadratic Functions Goals / “I can…”**

I can graph quadratic functions of the form y = ax + bx + c I can graph quadratic inequalities 2

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10.2 – Quadratic Functions 2 Yesterday we learned about y = ax and y = ax + c. The a changes the ????? and the c moves the parabola ?????. 2

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10.2 – Quadratic Functions 2 Today we look at y = ax + bx + c. The bx moves the parabola horizontally, changing the location of the line of symmetry.

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10.2 – Quadratic Functions 2 In an equation y = ax + bx + c, the line of symmetry can be found by the equation The x-coordinate of the vertex is

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**Finding the Line of Symmetry**

10.2 – Quadratic Functions Finding the Line of Symmetry When a quadratic function is in standard form For example… Find the line of symmetry of y = 3x2 – 18x + 7 y = ax2 + bx + c, The equation of the line of symmetry is Using the formula… Thus, the line of symmetry is x = 3 This is best read as … the opposite of b divided by the quantity of 2 times a.

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**10.2 – Quadratic Functions Finding the Vertex y = –2x2 + 8x –3**

We know the line of symmetry always goes through the vertex. STEP 1: Find the line of symmetry STEP 2: Plug the x – value into the original equation to find the y value. Thus, the line of symmetry gives us the x – coordinate of the vertex. y = –2(2)2 + 8(2) –3 y = –2(4)+ 8(2) –3 y = –8+ 16 –3 To find the y – coordinate of the vertex, we need to plug the x – value into the original equation. y = 5 Therefore, the vertex is (2 , 5)

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**A Quadratic Function in Standard Form**

10.2 – Quadratic Functions A Quadratic Function in Standard Form The standard form of a quadratic function is given by y = ax2 + bx + c There are 3 steps to graphing a parabola in standard form. MAKE A TABLE using x – values close to the line of symmetry. Plug in the line of symmetry (x – value) to obtain the y – value of the vertex. USE the equation STEP 1: Find the line of symmetry STEP 2: Find the vertex STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve.

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10.2 – Quadratic Functions Graph the equation y = -3x + 6x + 5 2

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**10.2 – Quadratic Functions Graphing Quadratic Inequalities**

Graphing a parabola with an inequality is similar to a line. We use solid and dashed lines and we shade above (greater) or below (less) the line.

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**Forms of Quadratic Inequalities y<ax2+bx+c y>ax2+bx+c y≤ax2+bx+c y≥ax2+bx+c**

Graphs will look like a parabola with a solid or dotted line and a shaded section. The graph could be shaded inside the parabola or outside.

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**10.2 – Quadratic Functions 1. Sketch the parabola y=ax2+bx+c**

(dotted line for < or >, solid line for ≤ or ≥) ** remember to use 5 points for the graph! 2. Choose a test point and see whether it is a solution of the inequality. 3. Shade the appropriate region. (if the point is a solution, shade where the point is, if it’s not a solution, shade the other region)

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**10.2 – Quadratic Functions Graph y ≤ x2 + 6x - 4**

Test point Graph y ≤ x2 + 6x - 4 * Opens up, solid line * Vertex: (-3,-13) Test Point: (0,0) 0≤02+6(0)-4 0≤-4 So, shade where the point is NOT!

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**10.2 – Quadratic Functions Graph: y > -x2 + 4x - 3**

Test Point Graph: y > -x2 + 4x - 3 * Opens down, dotted line. * Vertex: (2,1) x y 1 -3 * Test point (0,0) 0>-02+4(0)-3 0>-3

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**10.2 – Quadratic Functions 1) y ≥ x2 and 2) y ≤ -x2 + 2x + 4 SOLUTION!**

Graph both on the same coordinate plane. The place where the shadings overlap is the solution. Vertex of #1: (0,0) Other points: (-2,4), (-1,1), (1,1), (2,4) Vertex of #2: (1,5) Other points: (-1,1), (0,4), (2,4), (3,1) * Test point (1,0): doesn’t work in #1, works in #2.

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10.2 – Quadratic Functions Graph: y > x + x+ 1 2

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