Example: Solve 5x 2 + 6x + 1 = 0. Coefficients are: a = 5, b = 6, c = 1. Quadratic Formula: x = −b ± √ (b2 − 4ac) 2a. Put in a, b and c: x = −6 ± √ (62 − 4×5×1) 2×5. Solve: x = −6 ± √ (36− 20) 10. x = −6

The following quadratic function examples have their respective solution which details the process and reasoning used to arrive at the answer. Try to solve the exercises yourself before looking at the solution. EXAMPLE 1 Graph the quadratic function x 2 + 2. Solution EXAMPLE 2 Graph the quadratic function x 2 See more

## 6.4 Quadratic Functions and Their Graphs

The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows: f(x) = a (x - h) 2 + k The discriminant D of the quadratic equation: a x 2 + b x + c = 0 is given by D = b 2 - 4 a c If D = 0 , the quadratic equation a x 2 + b x

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Add them up and the height h at any time t is: h = 3 + 14t − 5t 2. And the ball will hit the ground when the height is zero: 3 + 14t − 5t 2 = 0. Which is a Quadratic Equation ! In Standard Form it looks like: −5t 2 + 14t + 3 = 0. It looks even

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Solve mathematic questions

To solve a math equation, you need to find the value of the variable that makes the equation true.

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Math can be a difficult subject for some students, but with a little patience and practice, it can be mastered.

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