Question

VERTEX FORM The vertex form of a quadratic function is \(y=a(x-h)^{2}+k\). Its graph is a parabola with vertex at \((\boldsymbol{h}, \boldsymbol{k})\). Use completing the square to write the quadratic function in vertex form. Then give the coordinates of the vertex of the graph of the function. $$y=-x^{2}-5 x+6$$

Short answer

The vertex form of the given quadratic function is \(y=-(x+ \frac{5}{2})^{2}+\frac{89}{4}\). The coordinates of the vertex of the graph are \((- \frac{5}{2}, \frac{89}{4})\).

Step-by-step solution

Rewrite the Quadratic Equation in Vertex Form

To rewrite the equation in vertex form, we have to complete the square. We group the x terms together, extract the coefficient of \(x^{2}\) as a common factor and complete the square inside the parenthesis: \[\begin{aligned} y &= -(x^{2}+5x)+6 \ &= - \left( x^{2} + 5x + \left(\frac{5}{2}\right)^{2} - \left(\frac{5}{2}\right)^{2}\right) + 6 \ &= - \left( x^2 + 5x + \left(\frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 \right) \ &= - \left( x + \frac{5}{2} \right)^2 + \left(\frac{5}{2}\right)^2 + 6 \ &= - \left( x + \frac{5}{2}\right)^2 + \frac{89}{4} . \end{aligned}\] This is the quadratic equation in vertex form.

Identify the Coordinates of the Vertex

From the standard form of the vertex form equation, we can tell that the vertex is (-h, k). In our case, the coefficient of x inside the parenthesis is +5/2, which means that h is -5/2. The value of k equals 89/4. Thus, the coordinates of the vertex are \((- \frac{5}{2}, \frac{89}{4})\).