Question

try 4: convert the quadratic $y = x^2 + 10x + 9$ into vertex form by finding the axis of symmetry.

Explanation:

Step1: Find axis of symmetry

The formula for the axis of symmetry of $y=ax^2+bx+c$ is $x=-\frac{b}{2a}$. Here $a=1$, $b=10$, so:
$x=-\frac{10}{2\times1}=-5$

Step2: Find vertex y-value

Substitute $x=-5$ into the original equation:
$y=(-5)^2 + 10(-5) + 9=25-50+9=-16$

Step3: Write vertex form

Vertex form is $y=a(x-h)^2+k$, where $(h,k)$ is the vertex. Here $a=1$, $h=-5$, $k=-16$:
$y=(x-(-5))^2 + (-16)=(x+5)^2-16$

Answer:

$y=(x+5)^2 - 16$