1Cademy - Finding Points on the Graph of $$f(x) = x^2

Learn Before

Zero Product Property

Example

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

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