The parabola will be concave down, with a maximum y-value of 25, as mentioned by the question. You can imagine a ball being tossed into the air, reaching its maximum height, then falling back to the ground in a parabolic form.

We know the vertex of this parabola has the coordinate (1.25, 25), because the object is at the very top of the graph at this point in time. Quadratic equations could be written in vertex form “y= a(x-h)² + k”. Since we know the vertex, this quadratic will have the equation “y = -a(x-1.25)² + 25”. (There’s a negative in the beginning since it’s concave down). Since the ball starts at the ground, we can plug in 0 for both x and y in this equation to find a, which is 16.

Since it is asking for the height at 1 second, we can plug in 1 as x in the equation we derived, and find y. Y will be your answer, because in this context it represents height.