what is the vertex of the graph of y=-(x-1)(x+4)?

Question

Exercise 10.3 pg 621

Answer 1

The vertex of this symmetrical parabola lies midway between the roots (zeroes), 1 and -4. The midway value of x is the average (1-4)/2=-3/2 or -1.5, when y=6.25.
Also, y=-x^2-3x+4 has a vertex when dy/dx=0,
so -2x-3=0 at the vertex, and x=-3/2. At x=-3/2 y=6.25. At x=-1 and -2, y=6, so the vertex (-1.5, 6.25) is a maximum.

Answer 2

Let y=f(x)=-(x-1)(x+4)
Simplify the equation: f(x)=-(x²+3x-4)

Convert the equation above into its vertex form:
f(x)=-{(x+3/2)²-(9/4)-4}=-(x+3/2)²+25/4

Let f1(x)=-x².   The graph of f1(x) is a parabola, being convex upwards (spills water), and symmetrical with respect to x-axis.   And the vertex is the origin,O(0,0).
So, f1(x) takes its maximum, y=0, at x=0.

 

Thus, f(x)is f1(x) shifted off 3/2 units leftwards, and then shifted off 25/4 units upwards.
So, the coordinates of the vertex of f(x) are: x=-3/2, y=25/4

 

The answer: The graph of y=f(x) is a parabola, being convex upwards, and
symmetrical with respect to line x=-3/2.
The coordinates of the vertex are x=-3/2 and y=25.   That is: the equation takes its maximum, y=25/4, at x=-3/2.