No, only the ball modeled by the table will pass through the hoop.
Practice makes perfect
Let's start by analyzing the first basketball thrown, which is modeled by the given quadratic function written in standard form.
y=-0.125x^2+1.84x+6
To determine if the first thrown will pass through the hoop or not, we must verify if the point (12,10) is on the parabola. To do that, we substitute x=12 and y=10 into the function, and check whether we obtain a true statement.
We obtained a false statement, which means the first ball will not pass through the hoop. To determine if the second throw will pass through the hoop, we have to find its equation using the given table.
x
y
2
10
4
12
10
12
Notice that the last two rows have the same y-coordinate. Since the axis of symmetry acts as a mirror, then it must be the midpoint between the corresponding x-coordinates.
x= x_1+x_2/2 ⇒ x=4+ 10/2=7
The axis of symmetry is the vertical line x=7. This means that the x-coordinate of the vertex is 7. We can write a partial equation for the parabola in vertex form.
y=a(x-7)^2 + k
To find the values of a and k, we will use the values in the table to set two equations.
Point ( 2, 10)
Point ( 4, 12)
Substitute
10=a( 2-7)^2 + k
12=a( 4-7)^2 + k
Simplify
10=25a + k
12=9a + k
With the obtained equations, we can write a system of equations.
10=25a + k & (I) 12=9a + k & (II)
Let's solve the system by using the Elimination Method.
Now that we have the values of a and k, we can write the equation that corresponds to the function modeled by the table.
y=- 1/8(x-7)^2+105/8
Finally, we can check whether the point (12,10) satisfies this equation.
