What functions do two graph points determine?

Task

Let $P = (0,1)$ and $Q = (1,2)$ in the $(x,y)$ plane.

1. Show that there is a unique linear function described by the equation $y = mx + b$ whose graph contains $P$ and $Q$: find $m$ and $b$.
2. Show that there is a unique exponential function described by the equation $y = ab^x$ whose graph contains $P$ and $Q$: find $a$ and $b$.
3. Show that there is more than one quadratic function described by the equation $y = ax^2 + bx +c$ whose graph contains $P$ and $Q$.

IM Commentary

The equation $y = mx + b$ has two constants: $m$, which gives the slope of the graph, and $b$, which gives the $y$-intercept. Asking for the graph of the equation to pass through two points $P$ and $Q$ determines these two numbers, and so determines the equation. One way to see this is to write two equations in $m$ and $b$ expressing the fact that the points $P$ and $Q$ lie on the graph, and solve this system of equations for $m$ and $b$.

The equation $y = ab^x$ also has two constants: $a$, which gives the $y$-intercept, and $b$, which determine the growth or decay factor. Two points $P$ and $Q$ either determine a unique exponential equation (as in this case) or are incompatible (see "Do two points always determine a linear/exponential equation?'').

The equation $y = ax^2 + bx + c$ has three constants, so asking for the graph of the equation to go through only two points does not determine the constants uniquely, as the solution shows.

Students may not be familiar with the term "unique" as used in the mathematical context. Therefore, teachers may need to explain that the term unique indicates one and only one. In a) and b) solutions must then identify an equation satisfying the described conditions and also examine whether it is the only such equation.

Solution

1. To find an equation for a linear function whose graph contains $(0,1)$ and $(1,2)$ observe that the $y$-intercept of the graph is 1 and the slope is $\frac{2-1}{1-0} = 1$ so, using the slope-intercept form $y = mx + b$ we have $$ y = x + 1. $$ The slope $m$ and the $y$-intercept $b$ were determined unambiguously by the two points $(0,1)$ and $(1,2)$ and so there is only one line containing the two points.
2. For the graph of an equation of the form $y = ab^x$ to pass through $(0,1)$ we must have $a = 1$ since $b^0= 1$. Once we know that $a = 1$ we can use the second point $(1,2)$ to find $b$: if the graph of $y = b^x$ contains $(1,2)$ then $$ 2 = b^1 = b. $$

   As in the linear example considered in part (a) above, the two constants $a$ and $b$ were determined completely in terms of the two points $(0,1)$ and $(1,2)$ and so the equation $$y = 2^x$$ is the only one of the form $y = ab^x$ whose graph passes through these two points.

3. For the point $(0,1)$ to lie on the graph of $y = ax^2 + bx + c$, the equation must be satisfied by $x = 0$, $y = 1$, so $$ 1 = a\cdot 0^2 + b \cdot 0 + c = c. $$ So $c = 1$. Then for the point $(1,2)$ to lie on the graph, we must have $$ 2 = a \cdot 1^2 + b \cdot 1 + 1 = a + b + 1. $$ So $a+b = 1$. There are infinitely many choices of $a$ and $b$ that satisfy this. One of them is $a = 1$, $b = 0$, which, along with $c = 1$, leads to the equation $$ y = x^2 + 1. $$ Another is $a = 2$, $b = -1$, which leads to $$ y = 2x^2 -x + 1. $$