Exercise

Given the quadratic polynomial in standard form $$6x^2+5x-6.$$

(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

$a=6, \quad b=5,\quad c=-6$

Step 2: Calculate discriminant

The discriminant is

\begin{align*} D&=b^2-4ac\\ &=(5)^2-4(6)(-6)\\ &= 25+144\\ &=169 \end{align*}

Step 3: Calculate x-coordinate of the vertex

The x-coordinate of the vertex is

\begin{align*} h &= -\frac{b}{2a}\\ &= -\frac{(5)}{2(6)}\\ &= -\frac{5}{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{(169)}{4(6)}\\ &= -\frac{169}{24} \end{align*}

Step 5: Conclusion

Thefore the vertex is

$\Big(-\frac{5}{12},-\frac{169}{24}\Big).$

(ii) Vertex form

$a(x-h)^2+k$ $h = -\frac{5}{12}, \quad k = -\frac{169}{24}$ \begin{align*} a(x-h)^2+k&=6\Big(x-\Big(-\frac{5}{12}\Big)\Big)^2+\Big(-\frac{169}{24}\Big)\\ &=6\Big(x+\frac{5}{12}\Big)^2-\frac{169}{24} \end{align*}

The vertex form is

$6\Big(x+\frac{5}{12}\Big)^2-\frac{169}{24}$

Step 1: Draw the vertex

Draw the vertex $$\big(-\frac{5}{12},-\frac{169}{24}\big).$$

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

Step 4: Draw point 2 units left

$$4a =4(6)=24$$ units up from vertex

Step 5: Draw point 2 units right

$$4a = 4(6) =24$$ units up from vertex