Question

Write the vertex form equation of each parabola given vertex (8, -1), y-intercept: -17?

Answer 1

This problem deals with obtaining the equation of a parabola given the vertex of the parabola and the y-intercept of the parabola. Parabola is one of the conic sections, other conic sections include hyperbola and ellipse. The general form of a parabola is given by $y = a{x^2} + bx + c$. Whereas the vertex form of the parabola is $\left( {y - k} \right) = a{\left( {x - h} \right)^2}$, where the vertex is $\left( {h,k} \right)$.

Complete step by step solution:
So here we are given with the vertex of the parabola and the y-intercept of the parabola, and we have to find the vertex form equation of the parabola from the given information.
Here given only the y-intercept of the parabola, which means there is no x-coordinate for this, so the point can be taken as (0, -17) which also passes through the parabola.
Consider the general vertex form equation of the parabola, as shown below:
$\Rightarrow \left( {y - k} \right) = a{\left( {x - h} \right)^2}$
Here the vertex $\left( {h,k} \right) = \left( {8, - 1} \right)$, hence substituting this point in the above equation.
$\Rightarrow \left( {y + 1} \right) = a{\left( {x - 8} \right)^2}$

Now to find the value of $a$, substitute the point (0, -17) which passes through the parabola, as shown below:
$\Rightarrow \left( { - 17 + 1} \right) = a{\left( {0 - 8} \right)^2}$
$\Rightarrow - 16 = 64a$
So the value of $a$, is given below:
$\therefore a = - \dfrac{1}{4}$
So the vertex form of parabola is given by:
$\Rightarrow \left( {y + 1} \right) = - \dfrac{1}{4}{\left( {x - 8} \right)^2}$.

Note: The general form of a parabola is given by $y = a{x^2} + bx + c$, it’s vertex is at some point which is not at origin and it has intercepts. Whereas the standard form of parabola is given by ${x^2} = 4ay$, where it has vertex at origin which is the point $\left( {0,0} \right)$.