## Exercise

Given the quadratic polynomial in standard form $$4x^2+4x-7$$

(i) Find the roots of the polynomial using the quadratic formula.

(ii) Draw the x-intercepts and the axis of symmetry.

(iii) Find the y-value of the vertex by evaluating the polynomial.

(iv) Graph the parabola using three points.

## Solution

#### Step 1: Write coefficients

The coefficients are

$a=4, \quad b=4,\quad c=-7$

#### Step 2: Calculate discriminant

The discriminant is

\begin{align*} D&=b^2-4ac\\ &=(4)^2-4(4)(-7)\\ &= 16+112\\ &=128 \end{align*}

#### Step 3: Apply quadratic formula

$\begin{gather*} -\frac{b}{2a} \pm \frac{\sqrt{D}}{2a}\\ = -\frac{(4)}{2(4)} \pm \frac{\sqrt{128}}{2(4)}\\ = - \frac{4}{8} \pm \frac{\sqrt{128}}{8}\\ =-\frac{1}{2} \pm \frac{\sqrt{64}\sqrt{2}}{8}\\ =-\frac{1}{2} \pm \frac{8\sqrt{2}}{8}\\ =-\frac{1}{2} \pm \sqrt{2} \end{gather*}$

Roots $$x_1 =-\frac{1}{2}-\sqrt{2}, \quad x_2 =-\frac{1}{2}+\sqrt{2}.$$

### (ii) x-intercepts and axis of symmetry

#### Step 1: x-intercepts

The x-intercepts are

$(x_1,0),\quad (x_2,0)$

which is

$\Big(-\frac{1}{2}-\sqrt{2},0\Big),\quad \Big(-\frac{1}{2}+\sqrt{2},0\Big).$

#### Step 2: Axis of symmetry

The axis of symmetry is the vertical line $$x=h$$ where

\begin{align*} h &= -\frac{b}{2a}\\ &= -\frac{(4)}{2(4)}\\ &=-\frac{4}{8}\\ &=-\frac{1}{2}. \end{align*}

Axis of symmetry

$x=-\frac{1}{2}.$

### (iii) y-value of vertex

Use $$4x^2+4x-7$$ with $$x=-\frac{1}{2}$$:

\begin{align*} k&=4\big(-\frac{1}{2}\big)^2+4\big(-\frac{1}{2}\big)-7\\ &=4\big(\frac{1}{4}\big) - \frac{4}{2} -7\\ &=\frac{4}{4}-2-7\\ &=1-2-7\\ &=-8. \end{align*}

The y-value of the vertex is $$k=-8$$.

Thus the vertex is at

$\big(-\frac{1}{2},-8\big).$

### (iv) Graph of parabola using three points

The x-intercepts $$(x_1,0)$$ and $$(x_2,0)$$ are

$\big(-\frac{1}{2}-\sqrt{2},0\big),\quad \big(-\frac{1}{2}+\sqrt{2},0\big)$

and the vertex $$(h,k)$$ is

$\big(-\frac{1}{2},-8\big).$