Test 2. Quadratic equation

Test 2. The quadratic equation \(f(x) = ax^2 + bx + c\) represents a parabola in the Cartesian coordinate system. Each point on the parabola has the same distance to the focus and directrix.
1. Draw the focus as a red cross. The standard form of the equation of a parabola with its vertex \((h,k)\) and a vertical axis of symmetry is \((x-h)^2 = 4p(y-k)\), where the vertex is at \((h,k)\) and \(|2p|\) is the distance from the focus to the directrix. \(|p|\) is the distance from the focus to the vertex or from the vertex to the directrix. The focus is at \((h,k+p)\) and the equation of the directrix is \(y = k-p\). If \(p > 0\), the parabola opens up as following. If \(p < 0\), the parabola opens down. 2. This test limits the orientation of the directrix to a horizontal line (slope is 0), which can be used as the basis for grade 10 algebra test review and benchmarking. jsxgraph supports the drawing of an arbitrary line in the form \(ax+by+c=0\). The parameters are specified in the order \([c,a,b]\).
3. Compute the distance from point A to the focus and the directrix to verify computations in the test.
4. Show the vertex of the parabola with its coordinate.
5. A parabola can have 0, 1, or 2 x-intercepts. The x-intercept has the y-coordinate of 0. If there is no x-intercept, the parabola is above or below the x-axis and therefore does not intersect the x-axis. If there is one x-intercept, the x-intercept is the vertex of the parabola. If there are 2 x-intercepts, the x-coordinates of the intercepts can be computed by setting \(y\) to 0, as following:
\(\leqalignno{
(x-h)^2 &= 4p(0-k) \cr
(x-h)^2 &= -4pk \cr
x-h &= \pm\sqrt{-4pk} \cr
x &= h\pm{2}\sqrt{-pk}}\)
Note that if \(p\) and \(k\) have same sign, there is no real solution since the parabola is open up and above the x-axis if \(p > 0\), and is open down and below the x-axis if \(p < 0\). 6. The y-intercept has the x-coordinate of 0. It can be computed by setting \(x\) to 0, as following:
\(\leqalignno{
(0-h)^2 &= 4p(y-k) \cr
h^2 &= 4p(y-k) \cr
\frac {h^2}{4p} &= y-k \cr
\frac {h^2}{4p} + k &= y \cr
y &= k+\frac {h^2}{4p}}\)
7. Show the quadratic form and the vertex form of the parabola.
Quadratic form: \(f(x) = ax^2 + bx + c\)
Vertex form: \((x-h)^2 = 4p(y-k)\)
8. Relationship between quadratic and vertex form.
The quadratic form of the parabola can be derived from the vertex formula by solving for \(y\).
\(\leqalignno{
(x-h)^2 &= 4p(y-k) \cr
x^2-2hx+h^2 &= 4py-4pk \cr
x^2-2hx+h^2+4pk &= 4py \cr
\frac {x^2-2hx+h^2+4pk}{4p} &= y \cr
\frac {x^2}{4p} – \frac {2h}{4p}x + \frac {h^2}{4p} + \frac {4pk}{4p} &= y \cr
y &= \frac {1}{4p}x^2 – \frac {h}{2p}x + \frac {h^2}{4p} + k}\)
In summary,
\(a = \dfrac {1}{4p}, b=-\dfrac {h}{2p}, c=\dfrac {h^2}{4p} + k\) (y-intercept).
9. Show the conic form of the parabola with values of coefficients.
10. Need to convert decimal coefficients \(a,b,c\) to fractional form or normalize to a LCM.
10. Show the tangent (derivative) of the parabola (regions of positive and negative slopes).
11. Show the integrals of the parabola.
12. jQuery UI to update the position of each component with precision.
13. Static labels (copyright, equations) are not at fixed positions and can be moved. There are two open-source software packages involved to generate the math chart: jsxgraph and mathjax. Mathjax supports standard math notation as shown on this page which fills probably all highschool math publication requirements.
14. Need a class to display the right angle square symbol and better support for polar coordinates. In javascript mathjax syntax is enclosed within a pair of $ while in regular HTML coding, use a pair of backslash and parentheses (left to open and right to close).