Chapter 4.7, Problem 41PPS

To determine

To write: the standard form of quadratic equation in the vertex form.

Answer to Problem 41PPS

The vertex form of quadratic equation is $y=3{\left(x-\frac{2}{3}\right)}^2-\frac{10}{3}$ and $\left(\frac{2}{3},-\frac{10}{3}\right)$ is the vertex and the axis of symmetry is $x=-48$ .

Explanation of Solution

Given information:

The quadratic equation is mentioned as $y+2=3x^2-4x$ .

Formula used:

The vertex form of the quadratic equation is in $\left(h,k\right)$ form.

The axis of symmetry always passes through vertex of parabola which is $x=-b^{2}a$ .

Calculation:

Consider the standard form of quadratic equation $y+2=3x^2-4x\Rightarrow y=3x^2-4x-2.$

Here we will divide the above quadratic equation by $3$ both sides.

Therefore the quadratic equation becomes $\frac{y}{3}=x^2-\frac{4}{3}x-\frac{2}{3}\Rightarrow \frac{y}{3}=\left(x^2-\frac{4}{3}x-\frac{2}{3}\right)$ .

Recall that for the vertex form of the quadratic equation is in $\left(h,k\right)$ form.

Therefore the vertex form of the quadratic equation is $f\left(x\right)=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

Therefore $\begin{array}{l}\frac{y}{3}=\left(x^2-\frac{4}{3}x-\frac{2}{3}\right)\Rightarrow \frac{y}{3}=x^2-2*x*\frac{2}{3}+{\left(\frac{2}{3}\right)}^2-{\left(\frac{2}{3}\right)}^2-\frac{2}{3}\Rightarrow \frac{y}{3}={\left(x-\frac{2}{3}\right)}^2-\frac{4}{9}-\frac{2}{3}\Rightarrow \frac{y}{3}={\left(x-\frac{2}{3}\right)}^2-\frac{10}{9}\\ \Rightarrow y=3{\left(x-\frac{2}{3}\right)}^2-\frac{10}{3}.\end{array}$

Therefore $y=3{\left(x-\frac{2}{3}\right)}^2-\frac{10}{3}$ is the vertex form of the quadratic equation where $\left(\frac{2}{3},-\frac{10}{3}\right)$ is the vertex.

Recall that the axis of symmetry always passes through vertex of parabola and the x coordinate of the vertex is the equation of the axis of symmetry.

Therefore the axis of symmetry is vertical line $x=-b^{2}a$ .

Therefore the axis of symmetry is $x=-b^{2}a\Rightarrow x=-{\left(-4\right)}^2*\left(3\right)=-48$ .