Question

Drag each tile to the correct box.

Answer 1

Answer: Please observe the attached image.

First image has one x-intercept, second image has 0 x-intercepts and third image of a quadratic equation has 2 x-intercepts.

Step-by-step explanation:

Given vertex at (2,0) and y-intercept (0,4).

We can see that vertex is at x-axis at x=2.

Therefore, x-intercept for first one is 1.

In second graph, we can see that parabola (graph) doesn't crossing out x-axis.

Therefore, there is no any x-intercept for second graph.

And for third one we are given a quadratic equation $x^2+x-2=0$

could be factored as (x-2)(x+1) =0

Applying zero product rule, we get

x-2=0  => x= 2

x+1=0   => x=-1.

Therefore, first image has one x-intercept, second image has 0 x-intercepts and third image of a quadratic equation has 2 x-intercepts.

Answer 2

For f(x), which has a vertex at (2,0), the y-intercept at (0,4) is above this vertex, so the parabola opens upward. This means that the vertex is the only point that touches the x-axis, so there is only 1 x-intercept.
For h(x), the graph does not have any x-intercepts.
For g(x) = x^2 + x - 2 = (x+2)(x-1), this intersects the x-axis at x = -2 and x = 1, so there are 2 x-intercepts.
From least to greatest: h(x), f(x), g(x).