Question

Given a parabola with zeros at x=0 and x=4 ,
which of the following MUST be true:
I. The parabola opens upward
II. The parabola has a line of symmetry
at x=2
III. The parabola intercepts the y-axis at
y=0
a. I only
b. III only
c. I and III only
d. II and III only
e. I, II, and III

Answer 1

The parabola could open either up or down:
\[y = (x-0)(x-4)=x^2-4x \textrm{ or } y = -(x-0)(x-4) = -x^2 + 4x\]
So it can't be \(\textrm{A, C,}\) or \(\textrm{E}\). Also, the axis of symmetry is always the midpoint of the two roots (or zeros, or x-intercepts). So \[x_{sym} = \dfrac{0+4}{2} = 2\]
means that \(\textrm{B}\) is false, leaving only \(\textrm{D}\).

And if you're curious about \(\textrm{III}\):
Zeros are points on a graph where \(y=0\). At \(x = 0\) (The \(y\)-axis), \(y\) also equals 0, meaning that the graph intercepts the y-axis at \(y=0\), or \((0,0)\).

Answer 2

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