Chapter 3, Problem 20RE

20. In Problems 20 and 21, find the quadratic function for which: Vertex is $(-1, 2)$; contains the point $\left(1,6\right)$

To determine

To calculate: The quadratic function from the given point and the vertex.

Answer to Problem 20RE

The quadratic function is $y=x^2+2x+3$.

Explanation of Solution

Given:

The given vertex is $(-1,2)$ and the give point is $(1,6)$.

Formula used:

The general formula of a quadratic equation is

$y=a{(x-h)}^2+k$, where $(h,k)$ is the vertex of the given equation and we can find the value of $a$ by substituting the given point in the general equation.

Calculation:

Let the equation of the parabola be $y=a{(x-h)}^2+k$.

The given vertex is $(h,k)=(-1,2)$.

Thus on substituting the vertex in the above equation, we get

$y=a{(x-(-1))}^2+2$

$\Rightarrow y=a{(x+1)}^2+2$

Now, we have to find the value of $a$.

On substituting the given point in the equation, we get

$6=a{(1+1)}^2+2$

$\Rightarrow 6-2=4a$

$\Rightarrow a=1$

Thus, the quadratic equation is

$y={(x+1)}^2+2$

$\Rightarrow y=x^2+2x+1+2$

$\Rightarrow y=x^2+2x+3$