Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$. Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$.
I know how to do it with the $x$ intercepts but I can't solve this.
 A: The vertex form of the equation of a parabola is 
$$f(x) = a(x - h)^2 + k$$
where $(h, k)$ is the vertex of the parabola.  In this case, we are given that $(h, k) = (5, 0)$.  Hence,
\begin{align*}
f(x) & = a(x - 5)^2 + 0\\
     & = a(x - 5)^2
\end{align*}
Since we also know the parabola passes through the point $(7, -2)$, we can solve for $a$ because we know that $f(7) = -2$.
\begin{align*}
a(7 - 5)^2 & = -2\\
a(2)^2 & = -2\\
4a & = -2\\
a & = -\frac{1}{2}
\end{align*}
Thus, the given parabola has equation 
$$f(x) = -\frac{1}{2}(x - 5)^2$$ 
A: Notice, there are two cases satisfying the given conditions 


*

*Parabola with vertex at $(5, 0)$ & arms opening in the positive x-direction given as $$y^2=4a(x-5)$$
since, parabola passes through the point $(7, -2)$ hence, setting $x=7$ & $y=-2$ in the equation, one should get $$(-2)^2=4a(7-5)\implies a=\frac{1}{2}$$ hence equation of parabola $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y^2=2(x-5)}}$$

*Downward parabola with vertex at $(5, 0)$ is given as $$(x-5)^2=-4ay$$
since, parabola passes through the point $(7, -2)$ hence, setting $x=7$ & $y=-2$ in the equation, one should get $$(7-5)^2=-4a(-2)\implies a=\frac{1}{2}$$ hence equation of parabola $$(x-5)^2=-2y$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y=-\frac{1}{2}(x-5)^2}}$$
A: With the ansatz $y=ax^2+bx+c$ we get:
\begin{gather*}
5=-\frac{b}{2a} \tag{Equation for vertex's $x$ coordinate} \\
c-\frac{b^2}{4a}=0 \tag{Equation for vertex's $y$ coordinate} \\
0=25a+5b+c \tag{Parabola passes through vertex} \\
-2=49a+7b+c \tag{Parabola passes through other point}
\end{gather*}
Equation 2 gives us that $\Delta=0$, hence $ax^2+bx+c=a(x-d)^2$ for some $d$. Equation 1 shows $d=5$, whence $y=a(x-5)^2$. This renders equation 3 obvious. Equation 4 finally gives $a$: $-2=a(7-5)^2$, or $-2=4a$, that is $a=-\frac12$. So we get the equation:
$$y=-\frac12(x-5)^2.$$
A: It is a parabola, hence we know it is of the form $y=ax^{2} + bx + c$
Recall that the vertex of a parabola is given by $(\frac{-b}{2a}, c-\frac{b^{2}}{4a})$
$\frac{-b}{2a} = 5$, $c-\frac{b^2}{4a} = 0$,
 and $-2=a*7^{2} + (7)b + c$
Solving this system, we get $a = -0.5$, $b=5$, $c=-12.5$ so 
$y=-0.5x^2 + 5x - 12.5$. 
Alternatively, if the question asks for vertex form, it's significantly easier. Recall that if $(p, q)$ is the vertex of a parabola, $y=a(x-p)^2 + q$. In this case, substitute in the given values and solve for $a$. 
