[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}[/tex]
now, let's expand the squared term to get the standard form of the quadratic.
[tex]\bf y=\cfrac{1}{9}(x-2)^2-1\implies y=\cfrac{1}{9}(x^2-4x+4)-1 \\\\\\ y=\cfrac{1}{9}x^2-\cfrac{4}{9}x+\cfrac{4}{9}-1\implies \stackrel{its~coefficient}{y=\stackrel{\downarrow }{\cfrac{1}{9}}x^2-\cfrac{4}{9}x-\cfrac{5}{9}}[/tex]
Answer:
Step-by-step explanation:
Remark
The general formula for the vertex of a parabola is
y = a(x - b) + c
What you know from the vertex is the formula describes b and c and the y intercept gives you the information to solve for a.
Solution
y = a(x - 2)^2 - 1 If you graph this, the vertex should be at (2,-1). I'll upload the graph at the end.
Now we need to solve for a
x = 0
y = 5
5 = a(0 - 2)^2 - 1
5 = a(4) - 1 Add 1 to both sides.
6 = a(4) Divide by 4
6/4 = a 4/4 Switch
a = 3/2
Answer
y = 3/2 (x - 2)^2 - 1
