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The Quadratic Formula Objective: To Solve Quadratic Equations using the Quadratic Formula, to write a polynomial function in standard form given its zeros The Quadratic Formula • The Quadratic Formula can be used to find the zeros of a quadratic function or the roots (solutions) of a quadratic equation • Remember that a quadratic equation has the form ax2 + bx + c = 0 • The Quadratic Formula is: b b2 4ac x 2a The Quadratic Formula • Solve: 2x2 – 3x + 1 = 0 • As this is already in standard form, we can see that: • a=2 b = -3 c=0 b b2 4ac x 2a (3) (3)2 4(2)(1) x 2(2) x 3 98 4 x 3 1 4 31 4 31 31 x or x 4 4 4 2 x or x The solution set is 4 4 {½ , 1} 1 x 1 or x 2 x The Quadratic Formula • Solve: 3x2 + 4x – 2 = 0 • As this is already in standard form, we can see that: • a=3 b=4 c = -2 b b2 4ac x 2a (4) (4)2 4(3)(2) x 2(3) x 4 16 24 6 4 40 x 6 4 4 10 6 4 2 10 x 6 2 10 x 3 x The solution set is 2 10 3 Finding the Equation of a Quadratic Function • Find the zeros of the function: • • • • x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x – 4 = 0 or x=4 f(x) = x2 – 3x – 4 x+1=0 x = -1 • If we know the zeros of a function, we can work backwards to determine the equation of the function. • Step 1: • Step 2: • Step 3: Use the zeros to determine the factors If given a value of “a”, write it in front of the factors Multiply to get the function into standard form **NOTE: THIS METHOD WORKS ONLY WHEN THE ZEROS ARE INTEGERS** Finding the Equation of a Quadratic Function • Find the function with zeros -2 and 5 • • • • Given the zeros, we know that x = -2 and x = 5 Working backward, we can find the factors (x + 2) and (x – 5) The function (in factored form) is f(x) = (x + 2)(x – 5) There is no given value of “a,” so the function in standard form is f(x) = x2 – 3x – 10 • Find the function with zeros 1 and 6 and a = 2 • • • • • x = 1 and x = 6 Factors (x – 1) and (x – 6) f(x) = 2(x – 1)(x – 6) f(x) = 2(x2 – 7x + 6) f(x) = 2x2 – 14x + 12 Finding the Equation of a Quadratic Function • Find the function with zero 4 and a = -3 • Because there is only one zero, and we know that a quadratic function must have two zeros, 4 must be a double zero • x=4 and x=4 • Factors: (x – 4) and (x – 4) • f(x) = -3(x – 4)2 • f(x) = -3(x2 – 8x + 16) • f(x) = -3x + 24x – 48 Finding the Equation of a Quadratic Function • Find the function with zeros -2 and 6 and y-intercept -24 • • • • • • • • • • • x = -2 and x = 6 Factors: (x + 2) and (x – 6) f(x) = a(x + 2)(x – 6) To find the value of a, we can substitute the y-intercept coordinates (0, -24) -24 = a(0 + 2)(0 – 6) -24 = a(2)(-6) -24 = -12a a=2 f(x) = 2(x + 2)(x – 6) f(x) = 2(x2 – 4x – 12) f(x) = 2x2 – 8x – 24 Finding the Equation of a Quadratic Function • Find the function with zeros 3/2 and -1 and a = -4 • As the zeros ARE NOT integers, we have to look at the “a” value at the end of the problem • • • • • • x = 3/2 and x = -1 2x = 3 and x = -1 Factors: (2x – 3) and (x + 1) f(x) = (2x – 3)(x + 1) f(x) = 2x2 – x – 3 The “a” value once the factors have been multiplied is 2, when it should be -4. We need to adjust the function to correct the “a” value. We do this by multiplying by -2. • f(x) = -2(2x2 – x – 3) • f(x) = -4x2 + 2x + 6 Finding the Equation of a Quadratic Function • Find the function with zeros -1 and -2 and one point at (-3, -4) • • • • • • • • • • • x = -1 and x = -2 Factors: (x + 1) and (x + 2) f(x) = a(x + 1)(x + 2) To find the value of a, we can substitute the point coordinates (-3, -4) -4 = a(-3 + 1)(-3 + 2) -4 = a(-2)(-1) -4 = 2a a = -2 f(x) = -2(x + 1)(x + 2) f(x) = -2(x2 + 3x + 2) f(x) = -2x2 – 6x – 4