Question

Number 1 only : find an equation of the form y-k=a(x-h)^2 with vertex (1, -5) and contains -4,3

Answer 1

The parabola equation in vertex form is given by:

`\begin{gathered} y=a(x-h)^2+k \\ h\colon\text{horizontal coordinate of the vertex} \\ k\colon\text{vertical coordinate of the vertex} \end{gathered}`

From the given question, we are provided with the following:

`\begin{gathered} \text{vertex (1,-5)} \\ \text{point (x,y)= (-4,3)} \end{gathered}`

We have to use the given parameters to find the parameter 'a' from the parabola equation.

Thus, we have:

`\begin{gathered} y=a(x-h)^2+k \\ 3=a(-4-1)^2+(-5)^{} \\ 3=a(-5)^2-5 \\ 3+5=25a \\ 25a=8 \\ a=(8)/(25) \end{gathered}`

Hence, the equation of the parabola is:

`\begin{gathered} y=(8)/(25)(x-1)^2+(-5) \\ y=(8)/(25)(x-1)^2-5 \end{gathered}`

The graph of the parabola on the xy-plane is shown below: