Can anyone help me solve the following problem tonight if possible. Many thanks in advance!!
I'm not good with quadratics and can't even find the function formula for these.
A quadratic function passes through the points (0, 4) and (6, 0).
Sketch two possible quadratic functions with each of the following properties.
a Two roots and a turning point when x = 4.
b Exactly one root.
Regards,
Paul
I'll do the first for you and you can do the other two but I'll provide a hint:
One 'trick' to building functions is to 'multiple by zero' for things you don't want at a particular spot/value. We want y equal to 0 when x is six. If we put in the zero and multiply everything else in our function by (x-6), then y will be zero when x=6, so
y(x) = 0 + (x-6) f(x) = (x-6) f(x)
for some f(x) and we have the point (6,0) on our function. For the other point we want y equal to 4 when x is zero. Using our other x value and noting that 0-6 is -6, if we let
f(x) = -(1/6) g(x) we would have, turning the x-6 around,
\(\displaystyle y(x) = \frac{6-x}{6} g(x)\)
Now all we need is for g(x) to be 4 when x=0, so put in the 4 and multiple everything else by zero when x=0
g(x) = 4 + h(x) (x-0) = 4 + h(x) x
and our y function becomes
y(x) = \(\displaystyle \frac{1}{6}\) (6-x) (4 + h(x) x)
Since we want a quadratic and x (x-6) is a quadratic, h(x) has to be a constant. Call that constant t and
y(x) = \(\displaystyle \frac{1}{6}\) (6-x) (4 + t x)
If you need to do so, put that in the standard form
y(x) = a x
2 + b x + c
That is, what are a, b, and c in terms of t?
(a) Two roots and a turning point when x = 4.
Choose two roots and, because they have to be the same distance from 4 (the turning point), we can call one 4-d and the other 4+d. Now we have a quadratic going through points (4-d,0) and (4+d,0).
(b) Exactly one root. Choose that root, call it r. Now we have a quadratic going through points (r,0) and (r,0).
