Question

use the graph to write the equation of the quadratic

1. graph of a parabola with x-intercepts at -2 and 3 (approx), y-intercept at -6, vertex near (0.5, -6.25) or similar, grid with x from -4 to 4 and y from -6 to 2

Explanation:

Step1: Identify x-intercepts

The graph intersects the x-axis at \( x = -2 \) and \( x = 3 \) (wait, looking again, the right intercept: from the grid, x=3? Wait no, let's check the grid. The x-axis: -4, -2, 0, 2, 4. The parabola crosses x at -2 and 3? Wait no, the right crossing: when x=3? Wait the grid lines: each square is 1 unit. So the left root is \( x = -2 \), right root: let's see, the parabola crosses x at \( x = 3 \)? Wait no, the graph: when x=3, y=0? Wait the grid: from x=2 to x=4, the parabola crosses x at x=3? Wait maybe I misread. Wait the equation of a parabola with roots \( r_1 \) and \( r_2 \) is \( y = a(x - r_1)(x - r_2) \). Let's find the roots. From the graph, the parabola crosses the x-axis at \( x = -2 \) and \( x = 3 \)? Wait no, let's check the y-intercept. The y-intercept is at (0, -6). Let's assume the roots are \( x = -2 \) and \( x = 3 \). Then the equation is \( y = a(x + 2)(x - 3) \). Now use the y-intercept (0, -6) to find a. Plug x=0, y=-6: \( -6 = a(0 + 2)(0 - 3) \) → \( -6 = a(2)(-3) \) → \( -6 = -6a \) → \( a = 1 \). Wait but let's check the vertex. Wait maybe the roots are \( x = -2 \) and \( x = 3 \)? Wait no, maybe I made a mistake. Wait the graph: when x=-2, y=0; when x=3, y=0? Wait the grid: x=3 is between 2 and 4. Alternatively, maybe the roots are \( x = -2 \) and \( x = 3 \), but let's re-examine. Wait the original graph: let's count the units. The left intersection is at x=-2 (since it's on the vertical line x=-2). The right intersection: let's see, from x=2 to x=4, the parabola crosses x at x=3? Wait the distance from x=0 to the right root: 3 units? Wait maybe the roots are \( x = -2 \) and \( x = 3 \), so the factored form is \( y = a(x + 2)(x - 3) \). Then using y-intercept (0, -6): \( -6 = a(2)(-3) \) → \( a = 1 \). So the equation is \( y = (x + 2)(x - 3) = x^2 - x - 6 \). Wait but let's check the vertex. The vertex of \( y = x^2 - x - 6 \) is at \( x = \frac{1}{2} \), \( y = (\frac{1}{2})^2 - \frac{1}{2} - 6 = \frac{1}{4} - \frac{2}{4} - \frac{24}{4} = -\frac{25}{4} = -6.25 \), but the graph's vertex is at (0.5, -6.25)? Wait the graph shows the vertex near (0, -6), but maybe my root is wrong. Wait maybe the roots are \( x = -2 \) and \( x = 3 \), but let's check again. Alternatively, maybe the roots are \( x = -2 \) and \( x = 3 \), and the equation is \( y = (x + 2)(x - 3) = x^2 - x - 6 \). Let's check x=0: y=-6, which matches the y-intercept. x=-2: y=0, correct. x=3: y=0, correct. So the equation is \( y = x^2 - x - 6 \). Wait but let's check another point. When x=1, y=1 - 1 - 6 = -6? No, that's not right. Wait maybe I messed up the roots. Wait the graph: let's look at the grid again. The parabola crosses the x-axis at x=-2 and x=3? Wait the right crossing: from the graph, when x=3, y=0? Let's see the grid lines: x=3 is between 2 and 4, and the parabola crosses there. The y-intercept is (0, -6). So the factored form is correct. Alternatively, maybe the roots are x=-2 and x=3, so the equation is \( y = (x + 2)(x - 3) = x^2 - x - 6 \).

Wait, maybe I made a mistake in the right root. Let's check the graph again. The user's graph: the x-axis has marks at -4, -2, 0, 2, 4. The parabola crosses the x-axis at x=-2 (left) and x=3 (right, since between 2 and 4, at x=3). The y-intercept is at (0, -6). So using factored form: \( y = a(x + 2)(x - 3) \). Plug in (0, -6): \( -6 = a(2)(-3) \) → \( -6 = -6a \) → \( a = 1 \). So the equation is \( y = (x + 2)(x - 3) = x^2 - x - 6 \).

Step2: Verify the equation

Let's check x=-2: \( y = (-2 + 2)(-2 - 3) = 0*(-5) = 0 \), correct. x=3: \( y = (3 + 2)(3 - 3) = 5*0 = 0 \), correct. x=0: \( y = (0 + 2)(0 - 3) = 2*(-3) = -6 \), correct. So the equation is \( y = x^2 - x - 6 \).

Answer:

The equation of the quadratic is \( y = x^2 - x - 6 \) (or in factored form \( y = (x + 2)(x - 3) \))