Quadratic Functions and their graphs Lesson 1.7

A quadratic function- f(x) = ax 2 + bx + c where a=/ 0 The graph of a quadratic function is called a parabola Two very special parts of a parabola: Vertex: The

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turning

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point It is either a maximum or

minimum.

Axis of symmetry: A vertical Line that passes through the Vertex.

The axis of symmetry can always be found by calculating: x = - (b) 2a Vertex: ( - b , f( - b) ) 2a 2a Discriminant: If b 2 – 4ac > 0 Parabola crosses x-axis twice. There will be two x-intercepts.

b 2 – 4ac = 0 Parabola is

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tangent

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to x-axis. There is only one x-intercept.

b 2 – 4ac < 0 Parabola never crosses the x-axis. No x-intercepts.

The axis of symmetry is a vertical line midway between the x-intercepts. Therefore it is the

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average

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of the x-intercepts.

Example: Find the intercepts , the axis of symmetry ,

and the vertex , of this parabola.

y = (x+4)(2x – 3) To find x-intercepts: Replace y with 0.

0 = (x+4)(2x – 3) Set each factor = 0 x + 4 = 0 & 2x – 3 = 0 x = - 4 x = 3/2 Since the x-intercepts have already been found Find the average of these to find the axis.

x = -4 + 3/2 = -2.5 = -1.25

Now

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2 2

Vertex = (-1.25, f(-1.25)) (-1.25,-15.125) Sketch the graph

Example: Sketch the graph of the parabola. Label the intercepts, the axis of symmetry, and the y = 2x vertex.

2 – 8x + 5 1 st find the Axis of Symmetry X = -(-8) = 8 = 2 2(2) 4 Now find the vertex: V = (2, f(2)) = (2,- 3) On this one, since b 2 – 4ac = 44 , Which is not a perfect square, this can Not be factored

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use the quadratic Formula to find x intercepts.

X = 0.78 & 3.22

If you remember ‘c’ is always the y-intercept Sooo

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y-int = 5.

Draw the ‘vertical axis’ at x = 2, Plot the vertex at (2,-3) Estimate the x-intercepts at 0.78 and 3.33, Plot the y-intercept at y = 5 and plot its symmetry point at (4,5) and sketch the parabola!

If the equation can be written in the form of : y = a(x – h) 2 + k --- vertex -- (h,k) axis of sym. -- x =

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h

’

Example: a) Find the vertex of the parabola y = - 2x 2 + 12x + 4 by completing the square.

y = -2x 2 + 12x + 4 = 0 1 st subtract 4 from both sides y – 4 = -2x 2 y – 4 = -2(x 2 + 12x – 6x ) factor out a – 2 from both ‘x’terms complete the square inside the parentheses b = -6/2 = (-3) 2 = 9

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y – 4 – 18 = -2(x 2 now add 9 inside the ( ) but add -2(9) or -18 to other side -6x +9 )

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change to this look y – 22 = -2(x – 3) 2 y = -2(x – 3) 2 + 22 Now add 22 back to the right side line up y = a(x – h) 2 + k y = a(x – h) 2 + k Identify h = 3, k = 22 so vertex = (h,k) = (3,22)

Example:

b) Find the

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x

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and

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y

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– intercepts.

y – intercept can be found from the given equation ‘c’ = y –intercept so y-intercept = 4 to find x-intercepts: let y = 0 and get: 0 = -2(x – 3) 2 + 22 - 22 = -2(x – 3) 2 11 = (x – 3) 2 + √11 = x – 3 3 + √11 = x Example: Find the equation of the quadratic function

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f

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with f(-1) = - 7 and a maximum value of f(2) = -1 f(-1) = - 7

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(-1,-7) A Maximum value at f(2) = -1 means (2,-1) is the vertex (h,k)

so using h = 2 and k = - 1 gives us this working format: y = a(x – 2) 2

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+ (-1) using the other point given (-1,-7) for x and y gives us: - 7 = a(-1 – 2) 2 - 1

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solve for ‘a’ + 1 + 1 - 6 = a(-3) 2 - 6 = 9a

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⅔ = a divide by 9 So putting it all together

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y = ⅔(x – 2) 2 - 1 b) Show that the function

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f

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no x-intercepts.

has The parabola must open downward since a = ⅔ and since the vertex is located below the x-axis at (2,-1) it cannot cross the x-axis!

Example: Where does the line y = 2x + 5 intersect the parabola y = 8 – x 2 Check out the solution for example 3 found on page 40!

(Show both the algebraic and graphical approach)

Example: Find an equation of the function whose graph is a parabola with x-intercepts

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1

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and

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4

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and a y-intercept of

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- 8

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Look over example 4 found on page 40.

Homework: page 40 CE #1-6 all; page 41 WE #1-25 left column