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**5.1 – Introduction to Quadratic Functions**

Objectives: Define, identify, and graph quadratic functions. Multiply linear binomials to produce a quadratic expression. Standard: E. Use equations to represent curves.

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**The Standard Form of a Quadratic Function is:**

(A quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.)

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A Quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.

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Ex 1a. Let f(x) = (2x – 5)(x - 2). Show that f represents a quadratic function. Identify a, b, and c when the function is written in the form f(x) = ax2 + bx + c. FOIL First – Outer – Inner – Last OR Distribute each term in the first set of parentheses to each term in the second set of parentheses! (2x – 5)(x – 2) = 2x2 – 4x – 5x + 10 2x2 – 9x + 10 a = 2, b = -9, c = 10

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p.278: #13-19 ODD

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**The graph of a quadratic function is called a parabola.**

Each parabola has an axis of symmetry, a line that divides the parabola into two parts that are mirror images of each other. The vertex of a parabola is either the lowest point on the graph or the highest point on the graph. Axis of Symmetry Vertex

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Ex 2a. Identify whether f(x) = -2x2 - 4x + 1 has a maximum value or a minimum value at the vertex. Then give the approximate coordinates of the vertex. First, graph the function: Next, find the maximum value of the parabola (2nd, Trace): Finally, max(-1, 3).

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**III. Minimum and Maximum Values**

Let f(x) = ax2 + bx + c, where a ≠ 0. The graph of f is a parabola. If a > 0, the parabola opens up and the vertex is the lowest point. The y-coordinate of the vertex is the minimum value of f. If a < 0, the parabola opens down and the vertex is the highest point. The y-coordinate of the vertex is the maximum value of f.

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**a. f(x) = x2 + x – 6 b. g(x) = 5 + 4x – x2 c. f(x) = 2x2 - 5x + 2**

d. g(x) = 7 - 6x - 2x2 Opens up, has minimum value Opens down, has maximum value Opens up, has minimum value Opens down, has maximum value

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**5.2 Solving Quadratic Equations (an introduction)**

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**Solve 5x2 – 19 = 231. Give exact solutions**

Solve 5x2 – 19 = Give exact solutions. Then give approximate solutions to the nearest hundredth.

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Ex 2a. Solve 4(x+2)2 = 49

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A rescue helicopter hovering 68 feet above a boat in distress drops a life raft. The height in feet of the raft above the water can be modeled by h(t) = -16t2 + 68, where t is the time in seconds after it is dropped. After how many seconds will the raft dropped from the helicopter hit the water? 68 ft When the raft hits the water, the height will = 0, so h(t) = 0: Since only positive values of time make sense, the answer is 2.1 seconds.

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**If ∆ABC is a right triangle with the right angle at C, then a2+ b2 = c2.**

When you apply the Pythagorean Theorem, use the principal square root because distance and length cannot be negative. a b c

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Homework: p 278 #20-26even, 27-32, 38-44even, 49 p 287 #24-30even, 36-42even, 50

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