## Precalculus (6th Edition) Blitzer

The equation of the required parabola is$f\left( x \right)=\frac{1}{2}{{\left( x+3 \right)}^{2}}-4$.
Putting in the value of the vertex, that is, $\left( -3,-4 \right)$ in the general form of parabola, we get: \begin{align} & f\left( x \right)=a{{\left( x-\left( -3 \right) \right)}^{2}}+\left( -4 \right) \\ & f\left( x \right)=a{{\left( x+3 \right)}^{2}}-4 \\ \end{align} Since the graph of $f\left( x \right)$ passes through the point $\left( 1,4 \right)$ , therefore the point $\left( 1,4 \right)$ satisfies the equation of $f\left( x \right)$. Thus, \begin{align} & 4=a{{\left( 1+3 \right)}^{2}}-4 \\ & =a{{\left( 4 \right)}^{2}}-4 \\ & =16a-4 \end{align} Simplifying, \begin{align} & 16a=8 \\ & a=\frac{1}{2} \end{align} Putting in the value of a in $f\left( x \right)=a{{\left( x+3 \right)}^{2}}-4$. Therefore, the equation of the parabola is $f\left( x \right)=\frac{1}{2}{{\left( x+3 \right)}^{2}}-4$.