QUADRATIC EQUATIONS AND FUNCTIONS CHAPTER 7 QUADRATIC EQUATIONS AND FUNCTIONS

7-1 Completing the Square

Completing the Square Transform the equation so that the constant term c is alone on the right side. If a, the coefficient of the second-degree term, is not equal to 1, then divide both sides by a. Add the square of half the coefficient of the first-degree term, (b/2a)2, to both sides (Completing the square)

Factor the left side as the square of a binomial. Completing the Square Factor the left side as the square of a binomial. Complete the solution using the fact that (x + q)2 = r is equivalent to x + q = ±r

Solve: a2 – 5a + 3= 0 Move the 3 to the other side a = 1 Complete the square, add (5/2)2 Factor Solve

Solve: x2 – 6x - 3= 0 Move the 3 to the other side a = 1 Complete the square, add (3)2 Factor Solve

Solve: 2y2 + 2y + 5 = 0 move the 5 to the other side divide both sides by 2 (a  1) Add (1/2)2 to both sides Factor Solve

Solve: 7x2 – 8x + 3 = 0 move the 3 to the other side divide both sides by 7 (a  1) Add (4/7)2 to both sides Factor Solve

7-2 Quadratic Formula

The Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0 (a  0) are given by the formula:

Solve 3x2 + x – 1 = 0 5y2 = 6y – 3 2x2 – 3x + 7 = 0

7-3 The Discriminant

Discriminant The discriminant is used to determine the nature of the roots of a quadratic equation and is equal to: D = b2 – 4ac

Discriminant Cases If D is positive, then the roots are real and unequal. If D is zero, then the roots are real and equal (double root) . negative, the roots are imaginary.

Find the Discriminant x2 + 6x – 2 = 0 3x2 – 4x√3 + 4 = 0 x2 – 6x + 10 = 0

Discriminant The discriminant also show you whether a quadratic equation with integral coefficients has rational roots. D = b2 – 4ac

Test for Rational Roots If a quadratic equation has integral coefficients and its discriminant is a perfect square, then the equation has rational roots.

Test for Rational Roots If the quadratic equation can be transformed into an equivalent equation that meets this test, then it has rational roots.

Find the Determinant and Identify the Nature of the Roots 3x2 - 7x + 5 = 0 2x2 - 13x + 15 = 0 x2 + 6x + 10 = 0

Equations in Quadratic Form 7-4 Equations in Quadratic Form

Quadratic Form An equation in quadratic form can be written as: a[f(x)]2 + b[(f(x)] + c = 0 where a  0 and f(x) is some function of x. It is helpful to replace f(x) with a single variable.

Example (3x – 2)2 – 5(3x -2) – 6 = 0 Let z = 3x – 2, then z2 – 5z – 6 = 0 Solve for z and then solve for x

Solve Using Quadratic Form (x + 2)2 – 5(x + 2) – 14 = 0 (3x + 4)2 + 6(3x + 4) – 16 = 0 x4 + 7x2 – 18 = 0

7-5 Graphing y – k = a(x- h)2

Parabola Parabola is the set of all points in the plane equidistant from a given line and a given point not on the line. Parabolas have an axis of symmetry (mirror image) either the x-axis or the y-axis. and

Parabola The point where the parabola crosses it axis is the vertex. The graph is a smooth curve.

Parabola The graph of an equation having the form y – k = a(x - h)2 has a vertex at (h, k) and its axis is the line x = h.

Graph y = x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0

Table of Values 1 -1 2 4 -2

Graph y = ½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0

Graph y = -½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0

Graph The graph of y = ax2 opens upward if a> 0 and downward if a< 0. The larger the absolute value of a is, the “narrower” the graph.

Graph y = ½(x-3)2 Use the form y – k = a(x – h)2 k= 0, h = 3, so the vertex is (3,0) and x = 3

Graph To graph y = a(x – h)2, slide the graph of y = ax2 horizontally h units. If h > 0, slide it to the right; if h < 0, slide it to the left. The graph has vertex (h, 0) and its axis is the line x = h.

Graph y – 3 = ½x2 Use the form y – k = a(x – h)2 k= 3, h = 0, so the vertex is (0,3) and x = 0

Graph y + 3 = ½x2 Use the form y – k = a(x – h)2 k= -3, h = 0, so the vertex is (0,-3) and x = 0

Graph To graph y – k = ax2, slide the graph of y = ax2 vertically k units. If k > 0, slide it upward; if k < 0, slide it downward. The graph has vertex (0, k) and its axis is the line x = 0.

7-6 Quadratic Functions

Completed square form: Quadratic Functions A function that can be written in either of two forms. General form: f(x) = ax2 + bx + c Completed square form: a(x-h)2 + k

Graph f(x) = 2(x – 3)2 + 1 y = 2(x – 3)2 + 1

Graph It’s a parabola with vertex (3,1) and axis x = 3

Graph f(x) = 3x2 – 6x + 1 y = 3x2 – 6x + 1

Graph y = 3x2 – 6x + 1 Rewrite the equation in the form : y – k = a(x – h)2 y -1 = 3(x2 – 2x) y – 1 + 3 = 3(x2 – 2x + 1) y + 2 = 3(x-1)2

Quadratic Functions Let f(x) = ax2 + bx + c, a0 If a < 0, f has a maximum value. If a > 0, f has a minimum value. and

Quadratic Functions The graph of f is a parabola. This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph.

Quadratic Functions Let f(x) = ax2 + bx + c, a0 This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph.

Example y = 1/2x2 + 3x – 7/4 Find the maximum or minimum value of f. Find the vertex of the graph of f

Example y = 1/2x2 + 3x – 7/4 a = ½, a > 0, then f has a minimum value. Minimum occurs when x = -b/2a x = -3/2*1/2 = -3

Example y = 1/2x2 + 3x – 7/4 y = ½(-3)2 + 3(-3) – 7/4 y = -25/4 So the minimum value of f is -25/4 and the vertex is (-3, -25/4)

Writing Quadratic Equations and Functions 7-7 Writing Quadratic Equations and Functions

and Theorem A quadratic equation with roots r1 and r2 is x2 – (r1 + r2)x + r1r2 = 0 or a[x2 – (r1 + r2)x + r1r2 ] = 0 and

Theorem The equation just given is equivalent to a[x2 – (sum of roots)x + product of roots ] = 0

Example Find a quadratic equation with roots (2 + i)/3 and (2 – i)/3 Sum of roots = (2 + i)/3 + (2 – i)/3 = 4/3 Product of roots = (2 + i)/3 (2 – i)/3= 5/9 x2 – 4/3x + 5/9 = 9x2 -12x + 5

Theorem If r1 and r2 are the roots of a quadratic equation ax2 + bx + c =0, then r1 + r2 = sum of roots = -b/a and r1r2 = product of roots = c/a

Example Find the roots of 2x2 + 9x + 5 = 0 r1 = -9 + 41 r2 = -9 - 41 4 4

Check 4 4 = -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 4 4 4 4 = -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 4 4 = 5/2 = c/a

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