Question

graph the equation $y = x^2 + 2x - 8$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex.
click to plot points. click points to delete them.

Explanation:

Step1: Find the roots (x-intercepts)

Set \( y = 0 \), so solve \( x^2 + 2x - 8 = 0 \). Factor the quadratic: \( (x + 4)(x - 2) = 0 \). Thus, \( x = -4 \) or \( x = 2 \). So the roots are \( (-4, 0) \) and \( (2, 0) \).

Step2: Find the vertex

For a quadratic \( y = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = 2 \), so \( x = -\frac{2}{2(1)} = -1 \). Substitute \( x = -1 \) into the equation: \( y = (-1)^2 + 2(-1) - 8 = 1 - 2 - 8 = -9 \). So the vertex is \( (-1, -9) \).

Step3: Find two more points

Let's choose \( x = 0 \): \( y = 0^2 + 2(0) - 8 = -8 \), so the point is \( (0, -8) \).
Let's choose \( x = 1 \): \( y = 1^2 + 2(1) - 8 = 1 + 2 - 8 = -5 \), so the point is \( (1, -5) \).

Now we have five points: \( (-4, 0) \), \( (2, 0) \), \( (-1, -9) \), \( (0, -8) \), \( (1, -5) \). Plot these points on the coordinate plane and draw a parabola through them.

Answer:

The five points to plot are \((-4, 0)\), \((2, 0)\), \((-1, -9)\), \((0, -8)\), and \((1, -5)\). The parabola opens upward (since \( a = 1 > 0 \)) and passes through these points.