b If a<0, the given graph is reflected in the x-axis.
A
a The expression f( p+q2) represents the minimum value of the function.
B
bIf a<0, does the previous answer change? Yes Explanation: If a<0, the expression f( p+q2) represents the maximum value of the function.
Practice makes perfect
a We are given the graph of the function f(x)=a(x-p)(x-q).
Since the quadratic function is written in intercept form, p and q are the x-intercepts of the graph. We see that the parabola opens upward, and therefore a> 0. The midpoint between the x-intercepts is p+q2, which corresponds to the axis of symmetry of the parabola, and also to the x-coordinate of the vertex.
From the above, we obtain the following expression for the vertex of the parabola.
Vertex: (p+q/2,f(p+q/2))
This means that the expression f( p+q2) represents the y-coordinate of the vertex of the parabola. In this case, since the parabola opens upwards, f( p+q2) is the minimum value of the function.
b This time we have to study the interpretation of f(p+q/2) when a< 0. This new condition will reflect the graph across the x-axis.
As in Part A, we can write the coordinates of the vertex of the parabola as follows.
Vertex: (p+q/2,f(p+q/2))
As before, f( p+q2) represents the y-coordinate of the vertex of the parabola. However, since this time the parabola opens upwards, f( p+q2) is the maximum value of the function.
Extra
Note
The only thing that changes is the sign of the y-coordinate. In Part A it was negative, while in Part B it is positive.
Conclusion
After solving parts A and B, we conclude that given a quadratic function written in intercept form f(x)=a(x-p)(x-q), its vertex is ( p+q2,f( p+q2)). If a>0, then f( p+q2) is the minimum value of the function. If a<0, then f( p+q2) is the maximum value.
