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We want to write the vertex form of the parabola whose vertex is $(1,9)$ and passes through point $P(4,12).$ To do so, let's first recall the vertex form of a quadratic function.
$y=a(x−h)_{2}+k $
In this form, $(h,k)$ is the vertex of the parabola. Since we are given that the vertex of our function is $(1,9),$ we have that $h=1$ and $k=9.$ We can use these values to partially write our equation.
$y=a(x−1)_{2}+9 $
Finally, to find the value of $a,$ we will use the fact that the function passes through $(4,12).$ We can substitute $4$ for $x$ and $12$ for $y$ in the above equation and solve for $a.$
Now that we know that $a=31 ,$ we can complete the equation of the function.
$y=31 (x−1)_{2}+9⇔y=31 (x−1)_{2}+9 $

$y=a(x−1)_{2}+9$

$12=a(4−1)_{2}+9$

Solve for $a$

SubTermSubtract term

$12=a⋅3_{2}+9$

CalcPowCalculate power

$12=a⋅9+9$

SubEqn$LHS−9=RHS−9$

$3=9a$

DivEqn$LHS/9=RHS/9$

$93 =a$

RearrangeEqnRearrange equation

$a=93 $

ReduceFrac$ba =b/3a/3 $

$a=31 $