Question

REASONING Provide a counterexample to the following statement. The vertex of a parabola is always the minimum of the graph.

Short answer

Counterexample is$y=-x^2+\mathbf{2}$ with vertex is maximum of the graph.

Step-by-step solution

Step 1. Define Vertex.

The axis of symmetry intersects a parabola at only one point is called vertex.

Step 2. Counterexample.

Consider the quadratic equation, $y=-x^2+2$. It has a vertex at $\left(0,2\right)$. Comparing it with the standard form of quadratic equation, $a=-1<0$. Therefore, the parabola opens downward and the vertex is the maximum of the graph.

Step 3. Graph.

Make a graph representing the above-mentioned equation.