To transform the quadratic equation [tex]\( y = x^2 + 6x + 10 \)[/tex] into its vertex form, we need to rewrite it as [tex]\( y = (x + h)^2 + k \)[/tex]. Here is the step-by-step process:
1. Start with the given equation:
[tex]\[
y = x^2 + 6x + 10
\][/tex]
2. Identify the coefficient of the [tex]\( x \)[/tex] term:
The equation is already in the form where the coefficient of [tex]\( x^2 \)[/tex] is 1, so we only focus on the coefficient of [tex]\( x \)[/tex], which in this case is 6.
3. To complete the square, calculate [tex]\( \left(\frac{b}{2}\right)^2 \)[/tex]:
Here, [tex]\( b = 6 \)[/tex]. So, [tex]\( \left(\frac{6}{2}\right)^2 = 3^2 = 9 \)[/tex].
4. Rewrite the equation by adding and subtracting this value inside the equation:
[tex]\[
y = x^2 + 6x + 9 - 9 + 10
\][/tex]
5. Group the perfect square trinomial and simplify the constants:
[tex]\[
y = (x + 3)^2 - 9 + 10
\][/tex]
6. Combine the constant terms:
[tex]\[
y = (x + 3)^2 + 1
\][/tex]
From the above steps, we can see that the vertex form of the quadratic equation is [tex]\( y = (x + 3)^2 + 1 \)[/tex].
Thus, the values of [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are:
[tex]\[
h = 3 \quad \text{and} \quad k = 1
\][/tex]
Therefore, in the vertex form [tex]\( y = (x + \square)^2 + \square \)[/tex], the values are:
[tex]\[
y = (x + 3)^2 + 1
\][/tex]
