Chapter 4.7, Problem 44PPS

To determine

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

Answer to Problem 44PPS

The vertex form of quadratic equation is $y={\left(x+0.7\right)}^2-1.69$ and $\left(-0.7,-1.69\right)$ is the vertex and the axis of symmetry is $x=-1.96$ .

Explanation of Solution

Given information:

The quadratic equation is mentioned as $y=x^2+1.4x-1.2$ .

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 $y=x^2+1.4x-1.2$ .

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 $y=x^2+1.4x-1.2\Rightarrow y=x^2+2*x*\frac{1.4}{2}+\frac{1.96}{4}-\frac{1.96}{4}-1.2\Rightarrow y={\left(x+\frac{1.4}{2}\right)}^2-0.49-1.2\Rightarrow y={\left(x+0.7\right)}^2-1.69$ .

Therefore $y={\left(x+0.7\right)}^2-1.69$ is the vertex form of the quadratic equation where $\left(-0.7,-1.69\right)$ 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 $x=-b^{2}a\Rightarrow x=-{\left(1.4\right)}^2*1=-1.96$ .