Find a quadratic rule from its turning point

A parabola has vertex (−1,3). Which form already puts the vertex in the right place?

Fit a parabola through V=(−1,3) and P=(1,−5). Change the candidate scale a. V stays fixed; Q marks the curve at x=1. Find the setting where Q meets P and the required and candidate outputs match

Which scale makes the curve pass through P?

A vertex leaves the scale unknown

Many parabolas share one vertex. An extra point whose x-coordinate differs from the vertex lets us solve for a. The extra point must fit the same rule

Read the supplied parameters

For vertex (2,−1), use h=2 and k=−1. If the graph also passes through (4,7), substitute x=4 and y=7. Do not treat the second point as another vertex

Solve for the scale

Evaluate the square, isolate the term containing a, then divide. Finally write the complete rule, including the known shifts

Rule

\(\displaystyle y=a(x-3)^2-2\)

A parabola has vertex (3,−2) and passes through (5,0). Find a.

Substitute x=5 and y=0, evaluate the square, then solve for a.

Which complete rule has vertex (3,−2) and scale a=½?

Rule

\(\displaystyle y=\frac12(x-3)^2-2\)

Use this constructed rule to find the output at x=2.

Square the bracket first; an exact fraction or decimal is acceptable

Check the original information

Substitute the extra point into your final rule. At x=h the output must be k, and for a≠0 this is the turning point. A matching ordinary point alone does not prove that a claimed vertex is a vertex

A proposed rule is y=2(x+2)²+1. What is its vertex?

When the extra point is not enough

If the extra point repeats the vertex, it gives no new equation for a. Every nonzero scale still works. If it has the same x-coordinate but a different y-coordinate, no function can satisfy both points

You know the vertex is (2,3), and the extra point is also (2,3). Is the quadratic rule uniquely determined?

A claimed vertex is (2,3), but another claimed point is (2,5). What follows?

A vertex is not an arbitrary point

Two ordinary points alone do not determine a unique quadratic. A stated vertex supplies extra symmetry information. Do not silently assume that one supplied point is the turning point

Two ordinary points are given, with no vertex or other information. What should you do?

Rule

\(\displaystyle y=ax^2+3\)

A canopy cross-section has vertex (0,3) and passes through (2,2), in metres. Find a.

Subtract the vertex height before dividing by 2²

Rule

\(\displaystyle y=-\frac14x^2+3\)

For that canopy, find its height at horizontal position x=1 metre

Use x=1 in the rule and keep the minus sign

Use the model only over its stated span

If the canopy model is specified only for −2≤x≤2 metres, do not interpret its distant parts as the actual roof. The mathematical parabola extends further than the physical object

Independent rule-construction task

Choose a vertex and an additional point at a different x-coordinate. Derive a nonzero a, write the rule, check both pieces of information and graph it digitally. Explain why the point determines a and what would fail if it repeated the vertex