Section 8.1 “Graph y = ax²”.

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Presentation transcript:

Section 8.1 “Graph y = ax²”

Quadratic Function Section 8.1 “Graph y = ax²” A nonlinear function that can be written in the standard form y = ax² + bx + c where a = 0. The graph of a quadratic function is in the form of a U-shape and called a parabola.

Identifying Characteristics of Quadratic Functions Vertex: (-1, -2) Axis of Symmetry: x = -1 Domain: All real numbers Range: y ≥ -2

Graphing: y = ax² y = x² y-axis x-axis Vertex- Axis of Symmetry- “Parent Quadratic Function” x y 1 1 2 4 3 9 -1 1 -2 4 x-axis -3 9 Vertex- the lowest or highest point of the parabola Axis of Symmetry- the line that divides a parabola into 2 symmetric parts

Section 8.2 “Graph y = ax² + c”

Graphing: y = ax² + c x-axis y-axis “Parent Quadratic Function” f(x) = x² g(x) = 3x² -3 Compare to the graph of the parent function f(x) = x2 g(x) = 3x²– 3 x y -3 1 2 9 x-axis -1 -2 9 g(x) is a vertical translation down 3 units and a vertical stretch by a factor of 3 y-axis

Zeros of a Function an x-value for which f(x) = 0. A zero of a function is an x-intercept of the graph of the function. To find the zeros of a function, graph the function and locate the x-intercepts. f(x) = -12x2 + 3 x = ½ and -½

Section 8.3 “Graph y = ax² + bx + c”

Section 8.3 “Graph y = ax² + bx + c” Properties of the Graph of a Quadratic Function y = ax² + bx + c is a parabola that: -opens up if a > 0 -opens down if a < 0 -is narrower than y = x² if the |a| > 1 -is wider than y = x² if the |a| < 1 -has an axis of symmetry @ x = -(b/2a) -has a vertex with an x-coordinate of -(b/2a) -has a y-intercept of c. So the point (0,c) is on the parabola

Graph y = 3x² - 6x + 2 Step 1: Determine if parabola opens up or down Step 2: Find and draw the axis of symmetry Step 3: Find and plot the vertex Step 4: Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values. x y 2 -1 11 Step 5: Reflect the points plotted over the axis of symmetry. Step 6: Draw a parabola through the plotted points. Minimum Value

Minimum and Maximum Values For y = ax² + bx + c, the y-coordinate of the vertex is the MINIMUM VALUE of the function if a > 0 or the MAXIMUM VALUE of the function if a < 0. y = ax² + bx + c; a > 0 y = -ax² + bx + c; a < 0 maximum minimum

Section 8.4 “Graphing f(x) = a(x – h)² + k”

Section 8.4 “Graphing f(x) = a(x – h)² + k” Vertex Form of a quadratic function is the form f(x) = a(x – h)2 + k, where a ≠ 0. The vertex of the graph of the function is (h, k) and the axis of symmetry is h.

Find the axis of symmetry and vertex of the graph of the function f(x) = a(x – h)2 + k Vertex: (h, k) Axis of Symmetry: x = h y = -6(x + 4)2 - 3 y = -4(x + 3)2 + 1 Axis of Symmetry: Axis of Symmetry: x = -4 x = -3 Vertex: Vertex: (-4, -3) (-3, 1)

y = 3(x – 2)² – 1 Graph: y = a(x – h)² + k. Compare to f(x) = x2 1 2 “Parent Quadratic Function” y = x² Axis of x = h symmetry: x = 2 Vertex: (h, k) (2, -1) x y 1 2 11 x-axis The graph of y is a vertical stretch by a factor of 3, a horizontal translation right 2 units and a vertical translation down 1 unit. y-axis

Vertex; (1,2); passes through (3, 10) Write a quadratic function in vertex form whose graph has the given vertex and passes through the given point. Vertex; (1,2); passes through (3, 10) y = a(x - h)2 + k y = a(x - 1)2 + 2 10 = a(3 - 1)2 + 2 10 = a(2)2 + 2 10 = 4a + 2 8 = 4a y = 2(x - 1)2 + 2 2 = a

Section 8.5 “Using Intercept Form”

Section 8.5 “Using Intercept Form” of a quadratic function is the form f(x) = a(x – p)(x – q), where a ≠ 0. The x-intercepts are p and q and the axis of symmetry is .

Find the axis of symmetry, vertex, and zeros of the function f(x) = a(x – p)(x – q) Intercepts: p, q Axis of Symmetry: x = y = -(x + 1)(x – 5) y = (x + 6)(x – 4) Zeros: Zeros: x = -6 & 4 x = -1 & 5 Axis of Symmetry: Axis of Symmetry: x = -1 x = 2 Vertex: (2, 9) Vertex: (-1,-25)

y = -5x2 + 5x y = -5x(x - 1) x-axis y-axis Graph: y = a(x – p)(x – q). Describe the domain and range. y = -5x2 + 5x y = -5x2 + 5x Write in INTERCEPT form y = -5x(x - 1) Intercepts: x = 0; x = 1 Axis of x = symmetry: x = 1/2 Vertex: y = -5x(x + 1) x-axis y = -5(1/2)(1/2 + 1) (1/2, 5/4) The domain of the function is ALL REAL NUMBERS. The range of the function is y ≤ 5/4. y-axis

Zeros of a Function f(x) = -12x2 + 3 an x-value for which f(x) = 0. A zero of a function is an x-intercept of the graph of the function. f(x) = -12x2 + 3 Graph (Standard Form) Intercept Form (Factor) To find the zeros of a function, graph the function and locate the x-intercepts. f(x) = -12x2 + 3 f(x) = -3(4x2 – 1) f(x) = -3(2x – 1)(2x + 1) x = ½ and -½

Section 8.6 “Comparing Linear, Exponential, and Quadratic Functions”

Section 8.6 “Comparing Linear, Exponential, and Quadratic Functions” y = mx + b y = ax2+bx+c y = abx

Comparing Linear, Exponential, and Quadratic Functions Differences and Ratios Linear Function the first differences are constant y = mx + b Exponential Function consecutive y-values are common ratios y = abx Quadratic Function the second differences are constant y = ax2+bx+c

y = a(x – p)(x – q) y = a(x - 4)(x - 8) 12 = a(2 - 4)(2 - 8) Tell whether the table represents a linear, exponential, or quadratic function. Then write the function. The second differences are constant. Therefore, the function is quadratic. y = a(x – p)(x – q) y = a(x - 4)(x - 8) 12 = a(2 - 4)(2 - 8) y = 1(x - 4)(x - 8) 12 = a(-2)(-6) y = x2 -12x + 32 12 = 12a 1 = a

are constant. Therefore, the function is linear. Tell whether the table represents a linear, exponential, or quadratic function. Then write the function. The first differences are constant. Therefore, the function is linear. Common difference (slope) Y-value when x = 0. y = mx + b y = 2x + b y = 2x + 1

y = abx y = (8)(b)x y = (8)(1/2)x Tell whether the table represents a linear, exponential, or quadratic function. Then write the function. Consecutive y-values have a common ratio. Therefore, the function is exponential. Y-value when x = 0. Commonratio. y = abx y = (8)(b)x y = (8)(1/2)x