# What is the y-value of the vertex of the function f(x) = -(x - 3)(x + 11)?

**Solution:**

f(x) = -(x - 3)(x + 11)

= -(x^{2} - 33 - 3x +11x)

= -x^{2} - 8x + 33 --- (1)

We know the graph of an equation y = ax^{2} + bx + c where a ≠ 0 is a parabola .

The parabola opens upwards if a > 0 and opens downwards if a < 0.

The vertex of the parabola is the point where the axis and parabola intersect.

Its x coordinate x = -b/2a and its y coordinate is found out by substituting x = -b/2a in the parabola equation.

The parabola given in the problem statement has a negative coefficient of x^{2} and hence it is a parabola which opens downwards.

Also for the equation, a = -1, b = -8 and c = 33.

Therefor the x-coordinate of the vertex is

x = -b/2a

= -(-8)/[2(-1)]

= - 4

Substituting the value of x = -4 in equation (1) we get,

Now y = -(-4)^{2} -8(-4) + 33

= -16 + 32 + 33 = 49

So the vertex point coordinates are (-4, 49) and the y value is 49.

## What is the y-value of the vertex of the function f(x) = -(x - 3)(x + 11)?

**Summary:**

The y-value of the vertex of the function f(x) = -(x - 3)(x + 11) is 49.

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