Determining the rule for a parabola
Theory
Determining the rule for a parabola means finding the quadratic function that fits given information. Use vertex form \(y=a(x-h)^2+k\) when the turning point is known, factored form \(y=a(x-p)(x-q)\) when the \(x\)-intercepts are known, and general form \(y=ax^2+bx+c\) with three points; then substitute a further point to solve for the unknown coefficients.
Every parabola is the graph of a quadratic function, and the same curve can be written in three equivalent forms. Vertex form \(y=a(x-h)^2+k\) shows the turning point \((h,k)\); factored form \(y=a(x-p)(x-q)\) shows the \(x\)-intercepts \(x=p\) and \(x=q\); and general form \(y=ax^2+bx+c\) shows the \(y\)-intercept \(c\). In each form the same coefficient \(a\) sets the direction and width of the curve.
To determine the rule, choose the form that matches the information given. A stated turning point points to vertex form; stated \(x\)-intercepts point to factored form; three arbitrary points call for general form. Choosing well reduces the work to solving for a single unknown \(a\) rather than three unknowns \(a\), \(b\) and \(c\).
Once the form is written with the known features substituted, use one further point on the curve to create an equation and solve for \(a\). For general form, substituting three points produces three simultaneous equations in \(a\), \(b\) and \(c\). A quadratic has three unknowns, so exactly three independent conditions are needed to pin the curve down.
Vertex form — the turning point is \((h,k)\):
Factored form — the \(x\)-intercepts are \(x=p\) and \(x=q\):
General form — the \(y\)-intercept is \(c\); three points give three equations:
How to determine the rule for a parabola
- Read the clues. A turning point suggests vertex form; \(x\)-intercepts suggest factored form; three unrelated points need general form.
- Write the form with knowns in place. Put the turning point into \(y=a(x-h)^2+k\), or the intercepts into \(y=a(x-p)(x-q)\), leaving only \(a\) unknown.
- Substitute the extra point. Replace \(x\) and \(y\) with the coordinates of a further point to get an equation in \(a\) (or, for general form, three equations in \(a,b,c\)).
- Solve. Solve for \(a\) (or solve the simultaneous equations for \(a\), \(b\) and \(c\)).
- State the rule. Write the equation; expand to \(y=ax^2+bx+c\) only if that form is asked for, and check with one of the given points.
| \(y\) | \(=\) | \(a(x-1)^{2}+2\) |
| \(10\) | \(=\) | \(a(3-1)^{2}+2\) |
| \(10\) | \(=\) | \(4a+2\) |
| \(a\) | \(=\) | \(2\) |
| \(y\) | \(=\) | \(a(x+1)(x-3)\) |
| \(5\) | \(=\) | \(a(4+1)(4-3)\) |
| \(5\) | \(=\) | \(5a\) |
| \(a\) | \(=\) | \(1\) |
| \(y\) | \(=\) | \(ax^{2}+bx+c\) |
| \(c\) | \(=\) | \(4\) |
| \((-1,5):\ a-b+4\) | \(=\) | \(5\ \Rightarrow\ a-b=1\) |
| \((1,7):\ a+b+4\) | \(=\) | \(7\ \Rightarrow\ a+b=3\) |
| \(2a\) | \(=\) | \(4\ \Rightarrow\ a=2,\ b=1\) |
| \((-b,c)\) | \(=\) | \((-3,2)\) |
| \(b\) | \(=\) | \(3,\quad c=2\) |
| \(4\) | \(=\) | \(a(-2+3)^{2}+2\) |
| \(4\) | \(=\) | \(a+2\ \Rightarrow\ a=2\) |
| \(f(x)\) | \(=\) | \(2(x^{2}+6x+9)+2\) |
| \(=\) | \(2x^{2}+12x+20\) |
Common pitfalls
Frequently asked questions
How do you find the rule of a parabola from its turning point?
Use vertex form \(y=a(x-h)^2+k\) with turning point \((h,k)\), then substitute one further point and solve for \(a\). A turning point of \((1,2)\) through \((3,10)\) gives \(10=4a+2\), so \(a=2\) and \(y=2(x-1)^2+2\).
How do you find the equation of a parabola from its x-intercepts?
Use factored form \(y=a(x-p)(x-q)\) with the intercepts \(p\) and \(q\), then substitute another point and solve for \(a\). For intercepts \(-1\) and \(3\) through \((4,5)\), \(5=5a\), so \(a=1\) and \(y=(x+1)(x-3)\).
How do you find a quadratic that passes through three points?
Use general form \(y=ax^2+bx+c\) and substitute each point to form three simultaneous equations in \(a\), \(b\) and \(c\). A point like \((0,4)\) gives \(c=4\) at once, simplifying the remaining two.
Which form of a quadratic should you use?
Match the form to the information: vertex form for a turning point, factored form for \(x\)-intercepts, general form for three unrelated points. A good choice leaves only \(a\) to find instead of all three coefficients.
What does the coefficient a tell you about a parabola?
It sets direction and width: \(a>0\) opens upward, \(a<0\) opens downward, and a larger \(|a|\) makes it narrower. You cannot assume \(a=1\) — it must come from a given point.
Can you find a parabola from only two points?
Not in general: a quadratic has three unknowns, so three independent conditions are needed. Two ordinary points are too few, but a turning point counts as extra information, so a turning point plus one more point is enough.