## Exercise

Given the quadratic polynomial in standard form $$6x^2+7x-3.$$

(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 three points.

## Solution

### (i) Vertex

#### Step 1: Write coefficients

The coefficients are

$a=6, \quad b=7,\quad c=-3$

#### Step 2: Calculate discriminant

The discriminant is

\begin{align*} D&=b^2-4ac\\ &=(7)^2-4(6)(-3)\\ &= 49+72\\ &=121 \end{align*}

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

The x-coordinate of the vertex is

\begin{align*} h &= -\frac{b}{2a}\\ &= -\frac{(7)}{2(6)}\\ &= -\frac{7}{12} \end{align*}

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

The y-coordinate of the vertex is

\begin{align*} k &= -\frac{D}{4a}\\ &= - \frac{(121)}{4(6)}\\ &= -\frac{121}{24} \end{align*}

#### Step 5: Conclusion

Thefore the vertex is

$\Big(-\frac{1}{4},-\frac{121}{24}\Big).$

### (ii) Vertex form

$a(x-h)^2+k$ $h = -\frac{7}{12}, \quad k = -\frac{121}{24}$ \begin{align*} a(x-h)^2+k&=6\Big(x-\Big(-\frac{7}{12}\Big)\Big)^2+\Big(-\frac{121}{24}\Big)\\ &=6\Big(x+\frac{7}{12}\Big)^2-\frac{121}{24} \end{align*}

The vertex form is

$6\Big(x+\frac{7}{12}\Big)^2-\frac{121}{24}$

### (iii) Graph of parabola

#### Step 1: Draw the vertex

Draw the vertex $$\Big(-\frac{7}{12},-\frac{121}{24}\Big).$$

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

$$a = 6$$ units up from vertex

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

$$a =6$$ units up from vertex