Determine a quadratic rule from exact information

Points where the arch meets the ground

\(\displaystyle (1, 0)\text{ and }(9, 0)\)

Point on the arch

\(\displaystyle (5, 32)\)

A parabolic arch meets the ground at x = 1 and x = 9 and passes through the point (5, 32). Which rule models the arch?

Change the vertical scale and compare the table rows and readouts

For every position of the slider, what stayed fixed and what changed?

Determine a quadratic rule

Use the known roots or vertex to reduce the unknown coefficients. A further point must have an input different from the roots or vertex input to determine the scale. For three distinct input values, substitute into y=ax²+bx+c and solve the three coefficient equations. Check a≠0: collinear data give a line, not a quadratic

1. 1Match the given feature Choose the form that shows the given roots or vertex, so only the scale is left to find \(\displaystyle x=1, 9\Rightarrow y=a(x-1)(x-9)\)
2. 2Use another point Use a point whose input is not a known root or the vertex input. Substitute into the chosen form, solve for the scale and check that it is nonzero \(\displaystyle 32=a(5-1)(5-9)\Rightarrow a=-2\)
3. 3Expand exactly Multiply out the brackets carefully when the general form is required \(\displaystyle y=-2(x-1)(x-9)=-2x^2+20x-18\)
4. 4Verify the given information Substitute every supplied point into the final rule. A further requested input can also be evaluated; do not call reused data an independent check \(\displaystyle y(5)=-2(5-1)(5-9)=32\)

Vertex

\(\displaystyle (4, 18)\)

Point on the roof

\(\displaystyle (1, 0)\)

A tunnel roof has vertex (4, 18) and passes through (1, 0), with coordinates in metres. Find the vertical scale a in the vertex form

Substitute x = 1 and y = 0 into the vertex form with h = 4 and k = 18, then solve for a

Vertex

\(\displaystyle (4, 18)\)

Point on the roof

\(\displaystyle (1, 0)\)

Vertical scale

\(\displaystyle a=-2\)

Using the same tunnel roof, find its rule in general form

Expand -2(x-4)^2 + 18 and collect like terms

x-intercepts

\(\displaystyle -2\text{ and }6\)

Which form displays the given roots directly and leaves only the scale to determine?

Find a rule from three points

Three distinct input values determine a unique polynomial of degree at most two. Non-collinear points give a genuine quadratic. Use general form, form three equations and eliminate coefficients

1. 1Substitute each point The point with input zero gives c directly. The other two points give independent equations in a and b. \(\displaystyle c=1,\quad a+b+c=0,\quad4a+2b+c=3\)
2. 2Eliminate one unknown Subtract twice the first reduced equation from the second to remove b. \(\displaystyle a+b=-1,\quad4a+2b=2\Rightarrow2a=4\)
3. 3Recover the coefficients Then a=2, b=−3 and c=1. The nonzero leading coefficient confirms a quadratic \(\displaystyle y=2x^2-3x+1\)
4. 4Check all supplied points Each point must lie on the recovered graph \(\displaystyle y(0)=1,\quad y(1)=0,\quad y(2)=3\)

Points

\(\displaystyle (0,-2),\ (1,1),\ (3,-5)\)

Determine the quadratic rule through (0,−2), (1,1) and (3,−5). Give the collected general form

Start with y=ax²+bx+c and substitute the input-zero point first

Use two equal-height points

For a nonconstant quadratic, two distinct inputs with the same output lie equally far from the axis. Their midpoint gives h, but does not by itself give the vertex height k. Use the third point to solve for a and k.

1. 1Locate the axis The equal-height points have midpoint input 2. \(\displaystyle h=\frac{-1+5}{2}=2\)
2. 2Use two different squared distances Substitute one equal-height point and the third point into the vertex form with axis at 2. \(\displaystyle 9a+k=5,\quad a+k=-3\)
3. 3Solve and expand Subtract to get a=1, then k=−4. The rule is quadratic and satisfies all three points \(\displaystyle 8a=8\Rightarrow y=(x-2)^2-4=x^2-4x\)

Points

\(\displaystyle (-2,7),\ (4,7),\ (0,-1)\)

A quadratic passes through (−2,7), (4,7) and (0,−1). Use the equal-height pair to determine its rule in collected general form

The equal-height inputs locate the axis, not the vertex height. Solve for both a and k