Question

How do you write a quadratic function in vertex form whose graph has the vertex $\left( { - 3,5} \right)$ and passes through the point $\left( {0, - 14} \right)$?

Answer 1

In this question we have to find the quadratic form whose vertex and point through which it passes through, we use the vertex form of the equation which is given by $y = a{\left( {x - h} \right)^2} + k$, where $\left( {h,k} \right)$ is the vertex, and now substituting the point through which the equation passes through in the vertex form we will get the value of $a$, and simplifying the equation we will get the required quadratic equation.

Complete step-by-step solution:
Given that the graph has the vertex $\left( { - 3,5} \right)$ and it passes through the point $\left( {0, - 14} \right)$,
We know that the vertex form of the graph is given by $y = a{\left( {x - h} \right)^2} + k$, where $\left( {h,k} \right)$ is the vertex,
Now given the vertex of the given function is $\left( { - 3,5} \right)$,
Comparing the vertex form we can say that,$h = - 3$ and $k = 5$,
Now substituting the values in the vertex form we get,
$\Rightarrow y = a{\left( {x - \left( { - 3} \right)} \right)^2} + 5$,
Now simplifying we get,
$\Rightarrow y = a{\left( {x + 3} \right)^2} + 5$,
So, the equation passes through the point $\left( {0, - 14} \right)$, now here $x = 0$ and $y = - 14$,
Substituting the values in the equation we get,
$\Rightarrow - 14 = a{\left( {0 + 3} \right)^2} + 5$,
Now simplifying we get,
$\Rightarrow - 14 = a{\left( 3 \right)^2} + 5$,
Now taking the square we get,
$\Rightarrow - 14 = a\left( 9 \right) + 5$,
So, now subtracting 5 to both sides we get,
$\Rightarrow - 14 - 5 = 9a + 5 - 5$,
Now eliminating the like terms,
$\Rightarrow 9a = - 19$,
Now dividing both sides with 9 we get,
$\Rightarrow \dfrac{{9a}}{9} = \dfrac{{ - 19}}{9}$,
Now simplifying we get,
$\Rightarrow a = \dfrac{{ - 19}}{9}$,
So, by substituting the values in the vertex form, i.e., $a = \dfrac{{ - 19}}{9}$,$h = - 3$ and $k = 5$ then the quadratic function will be,
\Rightarrow $$$$y = \dfrac{{ - 19}}{9}{\left( {x + 3} \right)^2} + 5,
Now taking L.C.M on the right hand side we get,
$\Rightarrow y = \dfrac{{ - 19{{\left( {x + 3} \right)}^2} + 45}}{9}$,
Now taking 9 to the left hand side we get,
$\Rightarrow 9y = - 19{\left( {x + 3} \right)^2} + 45$,
Now taking the square we get,
$\Rightarrow 9y = - 19\left( {{x^2} + 6x + 9} \right) + 45$,
Now multiplying the terms in the right hand side we get,
$\Rightarrow 9y = - 19{x^2} - 114x - 171 + 45$,
Now simplifying we get,
$\Rightarrow 9y = - 19{x^2} - 114x - 126$.
So, the quadratic function is $9y = - 19{x^2} - 114x - 126$.

$\therefore$The quadratic function in vertex form whose graph has the vertex $\left( { - 3,5} \right)$ and passes through the point $\left( {0, - 14} \right)$ will be equal to $9y = - 19{x^2} - 114x - 126$.

Note: Symmetric points are called the points which are equidistant from the axis of symmetry and lie on the $x$-axis and they are calculated as $x$-intercepts. To find the vertex of the parabola we can also make use of the standard form of the equation $y = a{x^2} + bx + c$, where the axis of symmetry or the $x$-coordinate is given by $x = \dfrac{{ - b}}{{2a}}$ and then we will find the value of $y$ from the equation of the parabola.