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The information given in this particular problem:

Axis is parallel to y-axis; graph passes through and $(4,11)$.$(3, 4)$ $(0,3)$

From this information, I know that it opens either upwards or downwards, therefore, having the form $y=ax^2+bx+c$.

Evaluating the function at the three given coordinates, I am left with three linear equations:

$c=3$; $1=9a+3b$; and $7=16a+4b$ I multiply the second equation by 4, the third one by 3, and subtract the second one from the third, to solve for a, but end up with an erroneous value, $a=17/12$. What did I do wrong?

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$16a+4b=7$ is a wrong equation. It should be $16a+4b=8$. –  Dan Shved Nov 21 '12 at 18:35

2 Answers 2

up vote 1 down vote accepted

You are right that $y = ax^2 + bx + c.$ This is because its axis is the $y$-axis. You have three unknowns, namely $a$, $b$ and $c$. Luckily your are given three points on the curve. We are going to use these points to create three simultaneous equations in $a$, $b$ and $c$.

The first point is $(x,y) = (4,11)$ and so we substitute $x=4$ and $y=11$ into $y = ax^2 + bx + c.$ The resulting equation is $11 = 16a + 4b + c.$ The second point is $(x,y) = (3,4)$ and so we substitute $x=3$ and $y=4$ into $y = ax^2 + bx + c.$ The resulting equation is $4 = 9a + 3b + c.$ The third point is $(x,y) = (0,3)$ and so we substitute $x=0$ and $y=3$ into $y = ax^2 + bx + c.$ The resulting equation is $3 = c.$ We need to solve the following system of simultaneous equations:

$$ \begin{array}{cCCcccc} 16a & + & 4b & + & c & = & 11 \ ,\\ 9a & + & 3b & + & c & = & 4 \ , \\ & & & & c & = & 3 \ . \end{array}$$

Clearly, $c=3$, and so we can substitute this into the first two equations. Rearranging gives:

$$ \begin{array}{ccccc} 16a & + & 4b & = & 8 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $$

The first equation has a common factor of $4$ and so we may divide both sides by $4$ to give:

$$ \begin{array}{ccccc} 4a & + & b & = & 2 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $$

Let's solve these equations by elimination. First, let's multiply both sides of the first equation by $3$:

$$ \begin{array}{ccccc} 12a & + & 3b & = & 6 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $$

Subtracting the second from the first gives $3a = 5$, which implies that $a=5/3.$ Next, we may substitute $a = 5/3$ into either equation; I will chose the first one. If $a=5/3$ and $12a + 3b = 6$ then $20 + 3b = 6$ and so $3b = -14$ and $b=-14/3.$ It follows that $a = 5/3,$ $b=-14/3$ and $c=3$. Using this:

$$y = \frac{5}{3}x^2 - \frac{14}{3}x + 3 \, . $$

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here is the system I get:

$16a + 4b + c = 11, 9a + 3b + c = 4, c = 3$.

Plugging in the last one:

$16a+4b = 8, 9a+3b = 1$

Divide left one by $4$ and right one by $3$:

$4a + b = 2, 3a + b = 1/3$

Subtract: $a = 2 - 1/3 = 5/3$ and now $b = 1/3 - 3a = 1/3 - 5 = -14/3$.

So the parabola must be $y = 5/3 x^2 - 14/3 x + 3$

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