hi
here's my question :
Construct an example of a polynomial so that its graph goes through the points
$\displaystyle \left ( -1,5 \right ) $ and $\displaystyle \left ( 3,-2 \right )$
thanks
If you want a quadratic, this is how to do it:
Pick one of the points as the vertex -- though it does matter which one you pick, because you get different answers otherwise, the problem is only asking for an example. Let's pick (3, -2).
The vertex form of the parabola is
$\displaystyle y = a(x - h)^2 + k$
Plug in the vertex for (h, k):
$\displaystyle y = a(x - 3)^2 - 2$
Plug in the other point for (x, y) to solve for a:
$\displaystyle \begin{aligned}
5 &= a(-1 - 3)^2 - 2 \\
5 &= (-4)^2 a - 2 \\
5 &= 16a - 2 \\
7 &= 16a \\
a &= \frac{7}{16}
\end{aligned}$
So the equation of the parabola would be
$\displaystyle y = \frac{7}{16}(x - 3)^2 - 2$
01
No, k can't be just anything.
To find a linear equation, you need to find the slope:
$\displaystyle \begin{aligned}
m &= \frac{y_2 - y^1}{y_2 - y^1} \\
&= \frac{-2 - 5}{3 - (-1)} \\
&= -\frac{7}{4}
\end{aligned}$
Use point-slope form:
$\displaystyle \begin{aligned}
y - y_1 &= m(x - x_1) \\
y - 5 &= -\frac{7}{4}(x + 1) \\
4y - 20 &= -7(x + 1) \\
4y - 20 &= -7x - 7 \\
7x + 4y - 20 &= -7 \\
7x + 4y &= 13 \\
\end{aligned}$
So you see, it does matter what the constant term is.
01