Answer:
[tex]y=\left[x-\left(-2\right)\right]^2+\left(-9\right)[/tex]Explanation:
Given the quadratic equation in standard form:
[tex]y=x^2+4x-5[/tex]1. Transpose the c-value to the left side of the equation.
[tex]y+5=x^2+4x[/tex]2. Complete the square of the expression on the right side of the equation to get a perfect square trinomial. Add the resulting term to both sides.
[tex]\begin{gathered} y+5+(\frac{4}{2})^2=x^2+4x+(\frac{4}{2})^2 \\ \implies y+5+(2)^2=x^2+4x+(2)^2 \end{gathered}[/tex]3. Add the numbers on the left and factor the trinomial on the right.
[tex]$ y+9=(x+2)^2 $[/tex]4. Transpose the number across to the right side to get the equation into the vertex form, y=a(x-h)²+k.
[tex]y=(x+2)^2-9[/tex]5. Make sure the addition and subtraction signs are correct to give the proper vertex form.
[tex]y=\left[x-\left(-2\right)\right]^2+\left(-9\right)[/tex]The vertex form of the given quadratic equation is:
[tex]y=\left[x-\left(-2\right)\right]^2+\left(-9\right)[/tex]