Vertex form of quadratic polynomial
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
\[6\Big(x+\frac{5}{12}\Big)^2-\frac{169}{24}\](iii) Graph of quadratic polynomial
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