Chapter 4.7, Problem 45PPS

To determine

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

Answer to Problem 45PPS

The vertex form of quadratic equation is $y={\left(x-\frac{1}{3}\right)}^2-3$ and $\left(\frac{1}{3},-3\right)$ is the vertex and the axis of symmetry is $x=-\frac{4}{9}$ .

Explanation of Solution

Given information:

The quadratic equation is mentioned as $y=x^2-\frac{2}{3}x+\frac{26}{9}$ .

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 $y=x^2-\frac{2}{3}x+\frac{26}{9}$ .

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 $y=x^2-\frac{2}{3}x+\frac{26}{9}\Rightarrow y=x^2-2*\frac{1}{3}*x+\frac{1}{9}-\frac{1}{9}-\frac{26}{9}\Rightarrow y={\left(x-\frac{1}{3}\right)}^2-\frac{27}{9}\Rightarrow y={\left(x-\frac{1}{3}\right)}^2-3$ .

Therefore $y={\left(x-\frac{1}{3}\right)}^2-3$ is the vertex form of the quadratic equation where $\left(\frac{1}{3},-3\right)$ 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 $x=-b^{2}a\Rightarrow x=-{\left(\frac{2}{3}\right)}^2*1=-\frac{4}{9}$ .