Question

Explain how you could write a quadratic function in a factored form that would have a vertex with an $x-$coordinate of $3$ and two distinct roots.

Answer 1

Step-1: Find a quadratic equation which is a parabola with a vertex at x-axis:

Given that the vertex lies on the $x-$coordinate of $3$.

The vertex lies on the axis of symmetry, so the axis of symmetry is $\mathrm{x}$.

The quadratic equation in vertex form is $\mathrm{y}={(\mathrm{x}-\mathrm{h})}^2+\mathrm{k}$.

where $(\mathrm{h},\mathrm{k})$ is the vertex of the parabola.

Substitute $\mathrm{h}=3$ in the above equation:

The quadratic equation of a parabola is $\mathrm{y}={(\mathrm{x}-3)}^2+\mathrm{k}..........\left(1\right)$.

Step-2: Find any two $x-$intercepts that are equal distance from the axis of symmetry $x$.

In order to find the two distinct roots the parabola must be upward facing with vertex below the $x-$axis or downward facing with vertex above the $x-$axis.

Substitute $\mathrm{k}=-1$in equation $\left(1\right)$:

$$\begin{gathered} \mathrm{y}={(\mathrm{x}-3)}^2-1 \\ \Rightarrow \mathrm{y}=({\mathrm{x}}^2-6\mathrm{x}+9)-1\left[\mathrm{using}{(\mathrm{a}-\mathrm{b})}^2={\mathrm{a}}^2-2\mathrm{bc}+{\mathrm{b}}^2\right] \\ \Rightarrow \mathrm{y}={\mathrm{x}}^2-6\mathrm{x}+8 \\ \Rightarrow \mathrm{y}={\mathrm{x}}^2-4\mathrm{x}-2\mathrm{x}+8 \\ \Rightarrow \mathrm{y}=\mathrm{x}(\mathrm{x}-4)-2(\mathrm{x}-4) \\ \mathrm{y}=(\mathrm{x}-2)(\mathrm{x}-4) \end{gathered}$$

Take $\mathrm{y}=\mathrm{x}-2$.

substitute $x=3$ in above equation:

$$\begin{gathered} \mathrm{y}=3-2 \\ =1 \end{gathered}$$

Take $\mathrm{y}=\mathrm{x}-4$.

substitute $x=3$ in above equation:

$$\begin{gathered} y=3-4 \\ =-1 \end{gathered}$$

Absolute value will be considered as distance cannot be negative .

Therefore, $2$ and $4$ are each $1$ unit distance from $x=3$.

Hence, quadratic function in a factored form that would have a vertex with an $x-$coordinate of $3$ and two distinct roots is $\mathrm{y}=(\mathrm{x}-2)(\mathrm{x}-4)$.