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The parabola and the quadratic function (higher level courseware)
In year \latex{ 9 } when having dealt with the quadratic functions we said that the graph of any quadratic function defined on the set of real numbers is a parabola, and we did prove it for the graph of the function \latex{\R \longrightarrow \R,} \latex{x\longmapsto x^{2}.} Using the coordinate geometry knowledge we have just gained we can also prove the general case.
THEOREM: The graph of any quadratic function defined on the set of real numbers is a parabola the axis ofwhich is parallel with the \latex{ y }-axis.
Proof
Let us consider the quadratic function \latex{ f }: \latex{\R\longrightarrow \R, f(x)=ax^{2}+bx+c,} and let us assume that \latex{a\gt 0.} The equation of the graph of\latex{ f } is \latex{y=ax^{2}+bx+c.} We transform the right-hand-side of this equation by completing the square:
\latex{y=ax^{2}+bx+c=a\times (x^{2}+\frac{b}{a}\times x+\frac{c}{a} )=a\times \left[(x+\frac{b}{2a} )^{2}-\frac{b^{2} }{4a^{2}+\frac{c}{a} } \right]=}
\latex{y=ax^{2}+bx+c=a\times (x^{2}+\frac{b}{a}\times x+\frac{c}{a} )=a\times \left[(x+\frac{b}{2a} )^{2}-\frac{b^{2} }{4a^{2}+\frac{c}{a} } \right]=}
\latex{=a\times (x+\frac{b}{2a} )^{2}-\frac{b^{2}-4ac }{4a}.}
This implies that
\latex{y+\frac{b^{2}-4ac }{4a}=a\times (x+\frac{b}{2a} )^{2}.}
By comparing it with the equation
\latex{y-v=\frac{1}{2p}\times (x-u)^{2}}
of the parabola “open upwards” with the vertex \latex{T(u;v)}, with the focal parameter \latex{ p }, we obtain that
\latex{u=-\frac{b}{2a}, v=-\frac{b^{2}-4ac }{4a}, p=\frac{1}{2a},} i.e.
\latex{f:} \latex{\R\longrightarrow \R,f(x)=ax^{2}+bx+c (a\gt 0)} is a parabola with the axis parallel with the \latex{ y }-axis, with the vertex \latex{ T } \latex{(-\frac{b}{2a};-\frac{b^{2}-4ac }{4a} )} and with the focal parameter \latex{\frac{1}{2a}.}
The proof is similar for the case \latex{a\lt 0.} Our statements relating to the parabola change only in the way that in this case the focal parameter of the parabola is \latex{-\frac{1}{2a}.}
With this we have proven the theorem.
The converse of theorem just proven is also true:
With this we have proven the theorem.
The converse of theorem just proven is also true:
THEOREM: There is exactly one quadratic function defined on the set of real numbers for any parabola with the axis parallel with the \latex{ y }-axis the graph of which is the parabola considered.
Proof
Let us consider a parabola “open upwards” with the axis parallel with the \latex{ y }-axis, with the vertex \latex{T(u;v)} and with the focal parameter \latex{ p }. Its equation is
\latex{y-v=\frac{1}{2p}\times (x-u)^{2}.}
Let us rearrange this equation so that the left-hand-side is equal to \latex{ y }, then let us rearrange the right-hand-side:
\latex{y=\frac{1}{2p}\times (x-u)^{2}+v=\frac{1}{2p}\times (x^{2}-2ux+u^{2} )+v=}
\latex{=\frac{1}{2p}\times x^{2}-\frac{u}{p}\times x+\frac{u^{2}+2pv }{2p}.}
The equation obtained is the equation of the graph of the quadratic function:
\latex{f: \R\longrightarrow \R, f(x)=\frac{1}{2p}\times x^{2}-\frac{u^{2}+2pv }{2p}.} In the case of a parabola “open downwards” the initial equation is
\latex{f: \R\longrightarrow \R, f(x)=\frac{1}{2p}\times x^{2}-\frac{u^{2}+2pv }{2p}.} In the case of a parabola “open downwards” the initial equation is
\latex{y-v=-\frac{1}{2p}\times (x-u) ^{2}}
the proof is done in a completely analogous.
We are summarising the two theorems proven above with a slightly different formulation.
Let \latex{ M } denote the set of quadratic functions defined on the set of real numbers, and let \latex{ P } denote the set of parabolas with the axis parallel with the \latex{ y }-axis.
We are summarising the two theorems proven above with a slightly different formulation.
Let \latex{ M } denote the set of quadratic functions defined on the set of real numbers, and let \latex{ P } denote the set of parabolas with the axis parallel with the \latex{ y }-axis.
THEOREM: One can give a one-to-one assignment between the elements of the sets \latex{ M } and \latex{ P }, i.e. there is one and only one parabola from the set \latex{ P } for every function from the set \latex{ M }, which is the graph of the function considered, and vice versa, there is one and only one function from the set \latex{ M } for every parabola from the set \latex{ P } the graph of which is the parabola considered.
Exercises
{{exercise_number}}. The two zeros of the quadratic function \latex{f:\R\longrightarrow \R} are
- \latex{0;4}
- \latex{-2;3;}
- \latex{-4;9;}
- \latex{-\sqrt{2};\sqrt{5};}
- \latex{-\frac{3}{4};\frac{5}{8}.}
Give the equation of the axis of symmetry of the graph.
{{exercise_number}}. The quadratic function \latex{f:\R\longrightarrow \R} takes its minimum at the place \latex{x_{0}}. Give the coordinates of the vertex of the graph, if
- \latex{x_{0}=1, f(x_{0} )=-2;}
- \latex{x_{0}=-3, f(x_{0} )=5;}
- \latex{x_{0}=8, f(x_{0} )=-5.}
{{exercise_number}}. Plot the graphs of the following quadratic functions defined on the set of real numbers in the Cartesian coordinate system.
- \latex{x\longmapsto x^{2}}
- \latex{x\longmapsto (\frac{x}{2} )^{2}-3}
- \latex{x\longmapsto -2(x-3)^{2}}
- \latex{x\longmapsto 2x^{2}+2x-12}
Give the focal parameter, the coordinates of the vertex and the focus, and the equation of the directrix of the parabola obtained in each case.
{{exercise_number}}. For which quadratic function defined on the set of real numbers is it true that the graph of it is the parabola with the following equation:
- \latex{6y-12=-(x+1)^{2};}
- \latex{y+3=-2\times (x-1)^{2}+5;}
- \latex{8y-2\times (x-5)^{2}=12?}
{{exercise_number}}. Set up the equation of the parabola with the axis parallel to the \latex{ y }-axis three points of which are \latex{A(-2;0), B(4;0), C(1;-5).} Give the quadratic function the graph of which is this parabola.
