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Express the following in vertex form

f(x)=x^(2)+6x-2

Express the following in vertex form $\newline$ $f(x)=x^{2}+6 x-2$

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Convert equations of conic sections from general to standard form

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Q. Express the following in vertex form

\newline

f(x)=x^{2}+6 x-2

Identify Coefficients: To express the quadratic function $f(x) = x^2 + 6x - 2$ in vertex form, we need to complete the square. $\newline$ Vertex form of a quadratic function is given by $f(x) = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. $\newline$ First, we need to identify the coefficient of $x^2$ , which is $1$ ( $a = 1$ ), and the coefficient of $x$ , which is $6$ ( $b = 6$ ).
Complete the Square: Next, we take half of the coefficient of $x$ , which is $6$ , and square it to complete the square. $(\frac{6}{2})^2 = 3^2 = 9.$ We will add and subtract this number inside the parentheses to complete the square.
Rewrite the Function: Now, we rewrite the function by adding and subtracting $9$ inside the parentheses: $f(x) = x^2 + 6x + 9 - 9 - 2$ .
Factor the Trinomial: We can now factor the perfect square trinomial $x^2 + 6x + 9$ as $(x + 3)^2$ : $f(x) = (x + 3)^2 - 9 - 2.$
Combine Constants: Finally, we combine the constants to simplify the expression: $f(x) = (x + 3)^2 - 11$ .

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