Example

Finding Points on the Graph of f(x)=x28x+3f(x) = x^2 - 8x + 3 when f(x)=4f(x) = -4

To find xx when f(x)=4f(x) = -4 for the polynomial function f(x)=x28x+3f(x) = x^2 - 8x + 3, substitute 4-4 for f(x)f(x) to form the equation 4=x28x+3-4 = x^2 - 8x + 3. Put the quadratic equation in standard form by adding 44 to both sides, resulting in x28x+7=0x^2 - 8x + 7 = 0. Factor the trinomial to obtain (x7)(x1)=0(x - 7)(x - 1) = 0. Apply the Zero Product Property to solve for xx, yielding x7=0x - 7 = 0 or x1=0x - 1 = 0, which gives the solutions x=7x = 7 and x=1x = 1. Consequently, the two points (7,4)(7, -4) and (1,4)(1, -4) lie on the graph of the function.

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Updated 2026-04-30

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