Example

Finding Points on the Graph of f(x)=x2+2x2f(x) = x^2 + 2x - 2 when f(x)=6f(x) = 6

To find xx when f(x)=6f(x) = 6 for the polynomial function f(x)=x2+2x2f(x) = x^2 + 2x - 2, substitute 66 for f(x)f(x) to create the equation 6=x2+2x26 = x^2 + 2x - 2. Put the quadratic equation in standard form by subtracting 66 from both sides, yielding x2+2x8=0x^2 + 2x - 8 = 0. Factor the trinomial to obtain (x+4)(x2)=0(x + 4)(x - 2) = 0. Apply the Zero Product Property to solve for xx, which gives x+4=0x + 4 = 0 or x2=0x - 2 = 0, resulting in x=4x = -4 and x=2x = 2. Because f(4)=6f(-4) = 6 and f(2)=6f(2) = 6, the two points that lie on the graph of the function are (4,6)(-4, 6) and (2,6)(2, 6).

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Updated 2026-04-30

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