Engage your students with effective distance learning resources. ACCESS RESOURCES>>

Finding Parabolas through Two Points

Alignments to Content Standards: F-LE.A.2


  1. Find all quadratic functions described by the equation $y = ax^2 + bx + c$ whose graph contains the two points $(1,0)$ and $(3,0)$. How are the graphs of these functions related to one another?
  2. Find all quadratic functions described by the equation $y = ax^2 + bx +c$ whose graph contains the two points $(1,1)$ and $(3,7)$. How are the graphs of these functions related to one another?

IM Commentary

The equation $$ f(x) = ax^2 + bx + c $$ has three constants $a,b,c$ which can take any real number value. This gives three degrees of freedom in specifying a quadratic function. Requiring the graph of $f$ to pass through a point $P_1$ puts one condition on $a,b,c$ while requiring the graph of $f$ to pass through $P_2$ puts a second condition on $a,b,c$. Assuming the $P_1$ and $P_2$ have distinct $x$-coordinates, this leaves one degree of freedom for the coefficients $a,b,c$ of a quadratic whose graph goes through $P_1$ and $P_2$. This extra degree of freedom is revealed as a "scaling factor'' in part a) while part b) is more subtle as the graphs do not share the same vertex.

The second solution to part (b) is quite sophisticated. While students should be able to follow the logic its primary use should be to challenge them to think creatively about the problem. The first solution to (b) is more grounded in classroom practice and should likely be the one emphasized for the majority of students.


Solution: System of Linear equations

  1. To find the quadratic functions $f(x) = ax^2 + bx + c$ whose graphs contain the points $(1,0)$ and $(3,0)$ we can evaluate $f$ at 1 and 0 to find \begin{eqnarray*} f(1) &=& a+b+c = 0,\\ f(3) &=& 9a+3b+c = 0. \end{eqnarray*} Solving the first equation for $c$ gives $c = -a- b$. Plugging this into the second equation gives $$ 9a+3b-a-b = 0 $$ or $8a+2b=0$ which is the same as $4a+b = 0$. We cannot determine $a$ or $b$ but for a given $a$ we find that $b = -4a$ and, plugging back into $c = -a -b$ we get that $c = -a-(-4a) = 3a$. So the general form of a quadratic whose graph contains the points $(1,0)$ and $(3,0)$ is $$ f_a(x) = ax^2 -4ax +3a. $$ Plugging in $x = 1,3$ we verify that $f_a(1) = f_a(3) = 0$. The list is complete because the argument above applies to any quadratic function $f$ with $f(1) = 0$ and $f(3) = 0$.

    Note that $f_a(x) = a(x-1)(x-3)$ so we have found the same system of parabolas as we did in the first solution.

  2. This argument also applies in the second setting, the only change being the values of the function at 1 and 3. \begin{eqnarray*} f(1) &=& a+b+c = 1,\\ f(3) &=& 9a+3b+c = 7. \end{eqnarray*} As above in the solution to a, these two equations are insufficient to solve for all three variables but will allow us to determine two of the variables in terms of the third. We find, solving these two equations for $c$: \begin{eqnarray*} c &=& 1-a-b,\\ c &=& 7-9a-3b. \end{eqnarray*} Setting these two equal gives $$ 8a+2b = 6. $$ Manipulating gives $b = 3 - 4a$ and substituting this back into one of the equations for $c$ gives $c = 3a-2$. So the general form for a quadratic polynomial satisfying $h(1) = 1$ and $h(3) = 7$ is $$ h(x) = ax^2 + (3-4a)x + (3a-2). $$

Solution: Solution 2

  1. If the graph of $y = f(x) = ax^2 + bx + c$ passes through $(1,0)$ and $(3,0)$ this means that $f(1) = 0$ and $f(3) = 0$. One such quadratic polynomial is $$ f(x) = (x-1)(x-3) = x^2 - 4x + 3. $$ Since multiplying the polynomial $f(x)$ by a real number $k$ will not influence the value of $f(x)$ at $1$ and $3$ we find that the graph of $$ f(x) = k(x-1)(x-3) = k(x^2 - 4x + 3) $$ also passes through $(1,0)$ and $(3,0)$. Below are graphs of this parabola for $k = 1,3,-1,$ and $-3$:


    To see that this list is complete, note that any quadratic $ax^2 + bx + c$ whose graph contains $(1,0)$ and $(3,0)$ must be divisible by $(x-1)$ and must be divisible by $(x-3)$. So the only quadratics whose graphs contain $(1,0)$ and $(3,0)$ are those of the form $$ f(x) = k(x-1)(x-3) $$ where $k$ is a real number. The graphs of these parabolas each have a vertex somewhere on the line $x = 2$. The constant $k$ determines where the vertex is located and this in turn determines whether the parabola opens upward or downward and how flat or steep it is.

  2. Here the given points $(1,1)$ and $(3,7)$ do not give any immediate insight into the choice of a quadratic. The $x$-coordinate grows by 2 moving from $(1,1)$ to $(3,7)$ while the $y$-coordinates grow by $6$. When $x$ grows from $1$ to $3$ the function $x^2$ grows by $8$ ($2$ more than $6$) while the function $x$ grows by $2$. So if we consider $x^2 - x$ it will have the appropriate growth as $x$ goes from $1$ to $3$. Unfortunately the values of $x^2 -x$ at $1$ and $3$ are $0$ and $6$ rather than $1$ and $7$. Adding one we find $$ f(x) = x^2 -x + 1 $$ and we see $f(1) = 1$ and $f(3) = 7$ as desired.

    We now look for a second quadratic function whose graph contains $(1,1)$ and $(3,7)$. To make sure we find a different one, we can look for one where the coefficient of $x^2$ is negative. The function $-x^2$ decreases by $8$ as $x$ increases from $1$ to $3$ while the function $x$ grows by $2$. To get growth of $6$ we can take $-x^2 + 7x$. Adjusting as above to get values of $1$ and $7$ when $x = 1,3$ gives $$ g(x) = -x^2 + 7x -5. $$

    It remains to use these two distinct quadratics to find the full list of quadratics whose graph contains $P_1$ and $P_2$. Observe that if $t$ is any real number then $$ h_t(x) = tf(x) + (1-t)g(x) $$ will satisfy \begin{eqnarray*} h_t(1) &=& tf(1) + (1-t)g(1) = t + (1-t) = 1\\ h_t(3) &=& tf(3) + (1-t)g(3) = 7t + 7(1-t) = 7. \end{eqnarray*}

    Below are graphs of $h_t(x)$ when $t = -1, 0, 1,2$:


    It remains to show that the set of functions $h_t(x)$ is a complete list of all quadratic functions whose graph contains the points $(1,1)$ and $(3,7)$. Let $q(x)$ be a quadratic function with $q(1) = 1$ and $q(3) = 7$. Notice that $$ h_t(0) = tf(0) + (1-t)g(0) = t(1) + (1-t)(-5) = 6t-5 $$ Choosing $t_0 = \frac{5 + q(0)}{6}$ we see that $$ h_{t_0}(0) = q(0). $$ We also know that $h_{t_0}(1) = q(1)$ and $h_{t_0}(3) = q(3)$. Hence the quadratic function $h_{t_0}(x) -q(x)$ takes the value 0 at the three real numbers 0,1,3. This is only possible if $$ h_{t_0}(x) - q(x) = 0 $$ and so the collection of quadratic polynomials $h_t(x)$ is a complete list of all quadratics polynomials taking the value 1 when $x = 1$ and 7 when $x = 3$.

    Unlike the family of quadratics in part a, here $x$-coordinate of the vertex of the parabolas is not the same. In fact, each real number $v$ gives exactly one parabola, whose vertex has $x$-coordinate equal to $v$, which passes through $(1,1)$ and $(3,7)$.