Parabolas – Vertex, Intercept, and Standard Forms

Parabolas can be expressed in a couple different forms. Each form has a different way to identify the vertex and the other parts of the parabola. Use the following notes to find the formulas for each form. Try out some practice problems!

Notes

$\text{Parabolas (Alternative Vertex Form)}$
$\text{Equation Vertex Form}$	$(x-h{)}^2=4p(y-k)$	$(y-k{)}^2=4p(x-h)$
$\text{Focus}$	$(h,k+p)$	$(h+p,k)$
$\text{Directrix}$	$y=k-p$	$x=h-p$
$\text{Opening Direction}$	$\text{up if }p>0,\text{ down if }p<0$

Problems

$\mathbf{\text{1)}}$ Graph $f(x)=(x-4{)}^2+2$

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$\mathbf{\text{2)}}$ Graph $f(x)=2(x+3{)}^2-1$

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$\mathbf{\text{3)}}$ Find the axis of symmetry of $y=2(x+3{)}^2-1$

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The axis of symmetry is $x=-3$

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$y=2(x+3{)}^2-1$

$a=2,h=-3,k=-1$

$\text{Axis of Symmetry: }x=h$

$\text{Axis of Symmetry: }x=-3$

$\mathbf{\text{4)}}$ Find the directrix of $y=2(x+3{)}^2-1$

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The directrix is $y=-\frac{9}{8}$

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$y=2(x+3{)}^2-1$

$a=2,h=-3,k=-1$

$\text{Directrix: }y=k-{\displaystyle \frac{1}{4a}}$

$\text{Directrix: }y=(-1)-{\displaystyle \frac{1}{4(2)}}$

$\text{Directrix: }y=-1-{\displaystyle \frac{1}{8}}$

$\text{Directrix: }y=-\frac{9}{8}$

$\mathbf{\text{5)}}$ Find the Focus of $y=2(x+3{)}^2-1$

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The focus is $(-3,-\frac{7}{8})$

Show Work

$y=2(x+3{)}^2-1$

$a=2,h=-3,k=-1$

$\text{Focus: }(h,k+\frac{1}{4a})$

$\text{Focus: }((-3),(-1)+\frac{1}{4(2)})$

$\text{Focus: }(-3,-1+\frac{1}{8})$

$\text{Focus: }(-3,-\frac{7}{8})$

$\mathbf{\text{6)}}$ Find the x-intercepts of $y=2(x+3{)}^2-1$

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The x-intercepts are $(-3-\frac{\sqrt{2}}{2},0)\text{ \& }(-3+\frac{\sqrt{2}}{2},0)$

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$y=2(x+3{)}^2-1$

$0=2(x+3{)}^2-1$

$0=2(x^2+6x+9)-1$

$0=2x^2+12x+18-1$

$0=2x^2+12x+17$

$a=2,b=12,c=17$

${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

${\displaystyle x=\frac{-(12)\pm \sqrt{(12{)}^2-4(2)(17)}}{2(2)}}$

${\displaystyle x=\frac{-(12)\pm \sqrt{144-136}}{4}}$

${\displaystyle x=\frac{-(12)\pm \sqrt{8}}{4}}$

${\displaystyle x=\frac{-(12)\pm 2\sqrt{2}}{4}}$

${\displaystyle x=\frac{-(6)\pm \sqrt{2}}{2}}$

$(-3-\frac{\sqrt{2}}{2},0)\text{ \& }(-3+\frac{\sqrt{2}}{2},0)$

$\mathbf{\text{7)}}$ Find the y-intercept of $y=2(x+3{)}^2-1$

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The y-intercept is $(0,17)$

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$y=2(x+3{)}^2-1$

$y=2((0)+3{)}^2-1$

$y=2(3{)}^2-1$

$y=2(9)-1$

$y=18-1$

$y=17$

The y-intercept is $(0,17)$

$\mathbf{\text{8)}}$ Use completing the square to rewrite the equation in standard form.$2x^2-12x-y+22=0$

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The equation is $y=2(x-3{)}^2+4$

$\mathbf{\text{9)}}$ $y=2x^2+k$ contains the points $(2,0)$ and $(3,a).$ What is the value of a?

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The answer is $a=10$

Show Work

$\text{Given }$

$y=2x^2+k$

$\text{Plug in }(2,0)$

$(0)=2(2{)}^2+k$

$(0)=2(4)+k$

$0=8+k$

$-8=k$

$\text{Update equation}$

$y=2x^2-8$

$\text{Plug in }(3,a)$

$(a)=2(3{)}^2-8$

$a=2(9)-8$

$a=18-8$

$a=10$

$\mathbf{\text{10)}}$ What is the equation of the parabola that has vertex $(-2,4)$ and contains the point $(0,8)$?

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The answer is $y={(x+2)}^2+4$

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$y=a(x-h{)}^2+k$

$h=-2,k=4,x=0,y=8$

$(8)=a{((0)-(-2))}^2+(4)$

$8=a{\left(2\right)}^2+4$

$8=4a+4$

$4=4a$

$1=a$

$y={(x+2)}^2+4$

$\mathbf{\text{11)}}$ What is the equation of the parabola that has vertex $(1,-4)$ and contains the point $(-1,12)$?

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The answer is $y=4{(x-1)}^2-4$

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$y=a(x-h{)}^2+k$

$h=1,k=-4,x=-1,y=12$

$(12)=a{((-1)-(1))}^2+(-4)$

$12=a{(-2)}^2-4$

$12=4a-4$

$16=4a$

$4=a$

$y=4{(x-1)}^2-4$

$\mathbf{\text{12)}}$ What is the equation of the parabola that has vertex $(5,8)$ and contains the point $(7,11)$?

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The answer is $y=\frac{3}{4}{(x-5)}^2+8$

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$y=a(x-h{)}^2+k$

$h=5,k=8,x=7,y=11$

$(11)=a{((7)-(5))}^2+(8)$

$11=a{\left(2\right)}^2+8$

$11=4a+8$

$3=4a$

${\displaystyle 34=a}$

$y=\frac{3}{4}{(x-5)}^2+8$

$\mathbf{\text{13)}}$ Express $y=x^2-6x$ in vertex form.

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The answer is $y={(x-3)}^2-9$

$\mathbf{\text{14)}}$ Find the axis of symmetry of $y=x^2-8x$.

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The answer is $x=4$

Show Work

$y=x^2-8x$

$a=1,b=-8,c=0$

$\text{Axis of Symmetry: }{\displaystyle x=\frac{-b}{2a}}$

$\text{Axis of Symmetry: }{\displaystyle x=\frac{-(-8)}{2(1)}}$

$\text{Axis of Symmetry: }{\displaystyle x=\frac{8}{2}}$

$\text{Axis of Symmetry: }{\displaystyle x=4}$

$\mathbf{\text{15)}}$ Identify the x-intercepts of $y=x^2-14x+54$.

Show Answer

There are no x-intercepts

See Related Pages$$

$\bullet \text{ Parabola Grapher}$
$\text{(Desmos.com)}$

$\bullet \text{ All Conic Section Notes}$
$$

$\bullet \text{ Equation of a Circle}$
$(x-h{)}^2+(y-k{)}^2=r^2\dots$

$\bullet \text{ Parabolas}$
$y=a(x-h{)}^2+k\dots$

$\bullet \text{ Axis of Symmetry}$
$x=-\frac{b}{2a}\dots$

$\bullet \text{ Ellipses}$
$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1\dots$

$\bullet \text{ Area of Ellipses}$
$\text{Area}=\pi ab\dots$

$\bullet \text{ Hyperbolas}$
$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1\dots$

$\bullet \text{ Conic Sections- Completing the Square}$
$x^2+8x+y^2-6y=11\Rightarrow (x+4{)}^2+(y-3{)}^2=36\dots$

$\bullet \text{ Conic Sections- Parametric Equations}$
$x=h+r\cos t$
$y=k+r\sin t\dots$

$\bullet \text{ Degenerate Conics}$
$x^2-y^2=0\dots$

$\bullet \text{ Andymath Homepage}$

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