Question

14.
the equation ( y = -(x - 1)(x + 5) ) defines a function.
plot a point on the grid below to show each x - intercept of this function.
plot a point on the grid below to show the y - intercept of this function.
plot a point on the grid below to show the minimum/maximum value of this function.
(grid with x - axis from - 10 to 10 and y - axis from - 10 to 10, with grid lines)

Explanation:

Step1: Find x - intercepts

To find the x - intercepts, we set \(y = 0\) in the equation \(y=-(x - 1)(x + 5)\). So, \(0=-(x - 1)(x + 5)\). This implies either \(x-1 = 0\) (so \(x = 1\)) or \(x + 5=0\) (so \(x=-5\)). So the x - intercepts are at the points \((1,0)\) and \((-5,0)\).

Step2: Find y - intercept

To find the y - intercept, we set \(x = 0\) in the equation. Substitute \(x = 0\) into \(y=-(x - 1)(x + 5)\): \(y=-(0 - 1)(0 + 5)=-(-1)(5)=5\). So the y - intercept is at the point \((0,5)\).

Step3: Find vertex (maximum/minimum)

First, expand the equation: \(y=-(x^{2}+5x - x - 5)=-(x^{2}+4x - 5)=-x^{2}-4x + 5\). For a quadratic function \(y = ax^{2}+bx + c\) (here \(a=-1\), \(b = - 4\), \(c = 5\)), the x - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). Substitute \(a=-1\) and \(b=-4\): \(x =-\frac{-4}{2\times(-1)}=\frac{4}{-2}=-2\). Now substitute \(x=-2\) into the equation \(y=-x^{2}-4x + 5\): \(y=-(-2)^{2}-4\times(-2)+5=-4 + 8+5 = 9\). Since \(a=-1<0\), the parabola opens downwards, so the vertex is a maximum point at \((-2,9)\).

Answer:

• x - intercepts: \((1,0)\) and \((-5,0)\)
• y - intercept: \((0,5)\)
• Maximum point: \((-2,9)\) (To plot on the grid, mark these points: \((1,0)\), \((-5,0)\), \((0,5)\), \((-2,9)\))