14.07.2022 - 23:41

Rewrite the \quadratic function in standard form and give the vertex. f(x) = x^2 – 6x +11 \y = boxed{space}(x – boxed{space})^2 + boxed{space} The vertex is at ( boxed{space}, boxed{space})

Question:

Rewrite the \quadratic function in standard form and give the vertex.

{eq}\displaystyle f(x) = x^2 – 6x +11 \y = boxed{space}(x – boxed{space})^2 + boxed{space} {/eq}

The vertex is at {eq}\displaystyle ( boxed{space}, boxed{space}) {/eq}

Answers (1)
  • Lanora
    April 11, 2023 в 14:05
    First, we complete the square to rewrite the function in standard form. {eq}begin{align*} f(x) &= x^2 - 6x +11 \ &= (x^2 - 6x + 9) + 2 \ &= (x - 3)^2 + 2 end{align*}{/eq} Thus, the \quadratic function in standard form is {eq}\displaystyle y = (x - 3)^2 + 2 {/eq}. From this form, we can see that the vertex is located at {eq}\displaystyle (3,2) {/eq}.
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