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

Hint: 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,
$$\Rightarrow {\displaystyle \frac{9a}{9}}={\displaystyle \frac{-19}{9}}$$,
Now simplifying we get,
$$\Rightarrow a={\displaystyle \frac{-19}{9}}$$,
So, by substituting the values in the vertex form, i.e., $$a={\displaystyle \frac{-19}{9}}$$,$$h=-3$$ and $$k=5$$ then the quadratic function will be,
$$\Rightarrow$$$$y={\displaystyle \frac{-19}{9}}{\left(x+3\right)}^2+5$$,
Now taking L.C.M on the right hand side we get,
$$\Rightarrow y={\displaystyle \frac{-19{\left(x+3\right)}^2+45}{9}}$$,
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,
$$\Rightarrow 9y=-19x^2-114x-171+45$$,
Now simplifying we get,
$$\Rightarrow 9y=-19x^2-114x-126$$.
So, the quadratic function is $$9y=-19x^2-114x-126$$.

$$\therefore$$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 $$9y=-19x^2-114x-126$$.

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 $$x={\displaystyle \frac{-b}{2a}}$$ and then we will find the value of $$y$$ from the equation of the parabola.