1.8: Parabolas

Math 3AC: Analytic Geometry and Calculus I · LibreTexts

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    From LibreTexts, under CC BY-NC-SA 4.0, as the source states it; its equations are written in TeX, which the PDF shows typeset. This page at LibreTexts.

    Quadratic functions are those of the form \(f(x)=ax^2+bx+c\) for some real numbers \(a,b,c\)

    When \(a=0\), this simplifies down to a line (which we have already studied fully), so assume \(a\ne 0\)
    We start with the graph of the standard parabola \(y=x^2\)

    clipboard_eac7a14703da0b53691adbdd353206fec.png

    (We say the point \(\textcolor{orange}{ (0,0) }\) is the vertex of the parabola.)

    A parabola is in vertex form when written \(y=a(x-h)^2+k\)
    For \(g(x)=x^2\), this graph is just a transformation of \(g(x)\): \(y=ag(x-h)+k\)
    \(y_{new}=a\cdot y_{old}+k\) and \(x_{old}=x_{new}-h\)

    \(x_{old}\) \(y_{old}\) \(x_{new}\) \(y_{new}\)
    -1 1 h-1 k+a
    0 0 h k
    1 1 h+1 k+a


    So when graph \(y=a(x-h)^2+k\), we can always plot three simple points:
    The vertex: \((h,k)\), and the two points \((h\pm 1, k+a)\)

    clipboard_ed31d7589fac58d19e4dcb951cbd9b196.png

    If you are given a quadratic, you can put it into vertex form by completing the square.
    (Remember that \( (x+h)^2=x^2+2hx+h^2\) and \((x-h)^2=x^2-2hx+h^2\) )

    So if we start off with \(x^2+bx+c\), then \(b\) should be doing the job of \(2h\):
    \(b=2h \longleftrightarrow \displaystyle{h=\frac{b}{2}}\)

    Therefore we should add and subtract \(\displaystyle{\left( \frac{b}{2}\right)^2}\) (which is the same as \(h^2\)) to our expression giving
    \(\displaystyle{x^2+bx+c=\left(x+\frac{b}{2}\right)^2+k}\) for some \(k\)

    Example \(\PageIndex{1}\)

    Write the following quadratics in vertex form and find the vertex \((h,k)\):

    (a) \(y=x^2+6x\) (b) \(y=x^2-2x+4\)

    Solution

    (a) Since \(b=6 \rightarrow h= 3\) we know \( h^2=\textcolor{red}{9}\)

    So we add and subtract it: \(y=(x^2+6x+ \textcolor{red}{9})- \textcolor{red}{9}\)

    And group up the first three as a perfect square to get \(y=(x+3)^2-9\)

    This is vertex form, and we can clearly see our vertex is at the point \((-3,-9)\)

    (b) Since \(b=-2\rightarrow h=-1\) we know \(h^2=\textcolor{purple}{1}\)

    So we add and subtract it: \(y=(x^2-2x+\textcolor{purple}{1})-\textcolor{purple}{1}+4\)

    And group up the first three as a perfect square to get \(y=(x-1)^2+3\)

    This is vertex form, and we see our vertex is the point \((1,3)\)

    If the leading term is \(ax^2\) instead of \(x^2\), your first step should be to factor out \(a\) from both \(x\) terms, then repeat the process above. (Just dont forget to distribute \(a\) when adding and subtracting \(h^2\))

    Example \(\PageIndex{2}\)

    Write the following quadratics in vertex form and find the vertex \((h,k)\):

    (a) \(y=2x^2-8x+5\) (b) \(y=-\dfrac{1}{2}x^2-6x-4\)

    Solution

    (a) First we factor out the \(\textcolor{green}{2}\) from the \(x\) terms: \(y=\textcolor{green}{2}(x^2-4x)+5\)

    Since \(b=-4 \rightarrow h= -2\) we know \( h^2=\textcolor{orange}{4}\)

    So we add and subtract it: \(y=\textcolor{green}{2}(x^2-4x+ \textcolor{orange}{4})- \textcolor{orange}{4}\cdot \textcolor{green}{2}+5\)

    Inside the parenthesis is a perfect square, so we write \(y=2(x-2)^2-3\)

    This is vertex form, and we can clearly see our vertex is at the point \((2,-3)\)

    (b) First we factor out the \(\textcolor{green}{-\dfrac{1}{2}}\) from the \(x\) terms: \(y=\textcolor{green}{-\dfrac{1}{2}}(x^2+12x)-4\)

    Since \(b=12\rightarrow h=6\) we know \(h^2=\textcolor{blue}{36}\)

    So we add and subtract it: \(y=\textcolor{green}{-\dfrac{1}{2}}(x^2+12x+\textcolor{blue}{36})-\textcolor{blue}{36}\cdot \left( \textcolor{green}{-\dfrac{1}{2}}\right)-4\)

    Inside the parenthesis is a perfect square, so we write \(y=-\dfrac{1}{2}(x+6)^2+14\)

    This is vertex form, and we see our vertex is the point \((-6,14)\)


    As we can see, completing the square is a great tool we can use to help turn quadratic functions from their general form \(f(x)=ax^2+bx+c\) into vertex form \(f(x)=a(x-h)^2+k\)
    This is definitely useful if you ever want to draw a parabola!

    Example \(\PageIndex{3}\)

    Sketch the graph of the function \(y=-2x^2+12x-14\)

    Solution

    First we put it into vertex form, so we factor out the \(-2\) from the \(x\) terms to get \(y=-2(x^2-6x)-14\)

    Then since \(b=-6\), we get \(h=-3\) so \(h^2=9\)

    We add and subtract it: \(y= {-2}(x^2-6x+ {9})- {9} \cdot \left( {-2} \right) -14\)

    Which simplifies into vertex form: \(y=-2(x-3)^2+4\)

    We have a vertex at \((3,4)\) and can sketch using the nearby points \((2,2)\) and \((4,2)\)

    clipboard_e946ff39bfc69e0169f76537977188a6c.png

    This page titled 1.8: Parabolas was last modified on Thu, 13 Mar 2025 20:04:47 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kenn Huber.