## Exercise

Given the quadratic polynomial in standard form $$2x^2+x-1.$$

(i) Find the vertex of the quadratic polynomial.

(ii) Find the vertex form of this quadratic polynomial.

(iii) Use the vertex form to graph this quadratic polynomial, using five points.

## Solution

### (i) Vertex

#### Step 1: Write coefficients

The coefficients are

$a=2, \quad b=1,\quad c=-1$

#### Step 2: Calculate discriminant

The discriminant is

\begin{align*} D&=b^2-4ac\\ &=(1)^2-4(2)(-1)\\ &= 1+8\\ &=9 \end{align*}

#### Step 3: Calculate x-coordinate of the vertex

The x-coordinate of the vertex is

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

#### Step 4: Calculate y-coordinate of the vertex

The y-coordinate of the vertex is

\begin{align*} k &= -\frac{D}{4a}\\ &= - \frac{(9)}{4(2)}\\ &= -\frac{9}{8} \end{align*}

#### Step 5: Conclusion

Thefore the vertex is

$\Big(-\frac{1}{4},-\frac{9}{8}\Big).$

### (ii) Vertex form

$a(x-h)^2+k$ $h = -\frac{1}{4}, \quad k = -\frac{9}{8}$ \begin{align*} a(x-h)^2+k&=2\Big(x-\Big(-\frac{1}{4}\Big)\Big)^2+\Big(-\frac{9}{8}\Big)\\ &=2\Big(x+\frac{1}{4}\Big)^2-\frac{9}{8} \end{align*}

The vertex form is

$2\Big(x+\frac{1}{4}\Big)^2-\frac{9}{8}$

### (iii) Graph of quadratic polynomial

#### Step 1: Draw the vertex

Draw the vertex $$\Big(-\frac{1}{4},-\frac{9}{8}\Big).$$

#### Step 2: Draw point 1 unit left

$$a = 2$$ units up from vertex

#### Step 3: Draw point 1 unit right

$$a =2$$ units up from vertex

#### Step 4: Draw point 1 unit right

$$4a =4(2)=8$$ units up from vertex

#### Step 5: Draw point 1 unit right

$$4a =4(2)=8$$ units up from vertex