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

\({\text{Parabolas (Alternative Vertex Form)}}\)
\({\text{Equation Vertex Form}}\)	\((x-h)^2=4p(y-k)\)	\((y-k)^2=4p(x-h)\)
\({\text{Focus}}\)	\((h,k+p)\)	\((h+p,k)\)
\({\text{Directrix}}\)	\(y=k-p\)	\(x=h-p\)
\({\text{Opening Direction}}\)	\(\text{up if } p\gt0, \text{ down if } p \lt 0\)

Problems

\(\textbf{1)}\) Graph \( f(x)=(x-4)^2+2 \)

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\(\textbf{2)}\) Graph \( f(x)=2(x+3)^2-1 \)

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\(\textbf{3)}\) 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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\(\,\,\,\,\,\,y=2(x+3)^2-1\)

\(\,\,\,\,\,\,a=2,\,\,\,h=-3,\,\,\,k=-1\)

\(\,\,\,\,\,\,\text{Axis of Symmetry: }x=h\)

\(\,\,\,\,\,\,\text{Axis of Symmetry: }x=-3\)

\(\textbf{4)}\) 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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\(\,\,\,\,\,\,y=2(x+3)^2-1\)

\(\,\,\,\,\,\,a=2,\,\,\,h=-3,\,\,\,k=-1\)

\(\,\,\,\,\,\,\text{Directrix: }y=k-\displaystyle\frac{1}{4a}\)

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

\(\,\,\,\,\,\,\text{Directrix: }y=-1-\displaystyle\frac{1}{8}\)

\(\,\,\,\,\,\,\text{Directrix: }y=-\frac{9}{8}\)

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

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

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\(\,\,\,\,\,\,y=2(x+3)^2-1\)

\(\,\,\,\,\,\,a=2,\,\,\,h=-3,\,\,\,k=-1\)

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

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

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

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

\(\textbf{6)}\) 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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\(\,\,\,\,\,\,y=2(x+3)^2-1\)

\(\,\,\,\,\,\,0=2(x+3)^2-1\)

\(\,\,\,\,\,\,0=2\left(x^2+6x+9\right)-1\)

\(\,\,\,\,\,\,0=2x^2+12x+18-1\)

\(\,\,\,\,\,\,0=2x^2+12x+17\)

\(\,\,\,\,\,\,a=2,\,\,\,b=12,\,\,\,c=17\)

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

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

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

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

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

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

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

\(\textbf{7)}\) 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 \( (0,17) \)

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

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The equation is \( y=2(x-3)^2+4 \)

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

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The answer is \( a=10\)

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\(\text{Given }\)

\(\,\,\,\,\,\,y=2x^2+k\)

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

\(\,\,\,\,\,\,(0)=2(2)^2+k\)

\(\,\,\,\,\,\,(0)=2(4)+k\)

\(\,\,\,\,\,\,0=8+k\)

\(\,\,\,\,\,\,-8=k\)

\(\text{Update equation}\)

\(\,\,\,\,\,\,y=2x^2-8\)

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

\(\,\,\,\,\,\,(a)=2(3)^2-8\)

\(\,\,\,\,\,\,a=2(9)-8\)

\(\,\,\,\,\,\,a=18-8\)

\(\,\,\,\,\,\,a=10\)

\(\textbf{10)}\) 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 \( y=\left(x+2\right)^2+4\)

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\(\,\,\,\,\,\,y=a(x-h)^2+k\)

\(\,\,\,\,\,\,h=-2,\,\,\,k=4,\,\,\,x=0,\,\,\,y=8\)

\(\,\,\,\,\,\,(8)=a\left((0)-(-2)\right)^2+(4)\)

\(\,\,\,\,\,\,8=a\left(2\right)^2+4\)

\(\,\,\,\,\,\,8=4a+4\)

\(\,\,\,\,\,\,4=4a\)

\(\,\,\,\,\,\,1=a\)

\(\,\,\,\,\,\,y=\left(x+2\right)^2+4\)

\(\textbf{11)}\) 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 \( y=4\left(x-1\right)^2-4\)

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\(\,\,\,\,\,\,y=a(x-h)^2+k\)

\(\,\,\,\,\,\,h=1,\,\,\,k=-4,\,\,\,x=-1,\,\,\,y=12\)

\(\,\,\,\,\,\,(12)=a\left((-1)-(1)\right)^2+(-4)\)

\(\,\,\,\,\,\,12=a\left(-2\right)^2-4\)

\(\,\,\,\,\,\,12=4a-4\)

\(\,\,\,\,\,\,16=4a\)

\(\,\,\,\,\,\,4=a\)

\(\,\,\,\,\,\,y=4\left(x-1\right)^2-4\)

\(\textbf{12)}\) 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 \( y=\frac{3}{4}\left(x-5\right)^2+8\)

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\(\,\,\,\,\,\,y=a(x-h)^2+k\)

\(\,\,\,\,\,\,h=5,\,\,\,k=8,\,\,\,x=7,\,\,\,y=11\)

\(\,\,\,\,\,\,(11)=a\left((7)-(5)\right)^2+(8)\)

\(\,\,\,\,\,\,11=a\left(2\right)^2+8\)

\(\,\,\,\,\,\,11=4a+8\)

\(\,\,\,\,\,\,3=4a\)

\(\,\,\,\,\,\,\displaystyle{3}{4}=a\)

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

\(\textbf{13)}\) Express \(y=x^2-6x \) in vertex form.

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The answer is \( y=\left(x-3\right)^2-9\)

\(\textbf{14)}\) Find the axis of symmetry of \(y=x^2-8x \).

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The answer is \( x=4\)

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\(\,\,\,\,\,\,y=x^2-8x\)

\(\,\,\,\,\,\,a=1,\,\,\,b=-8,\,\,\,c=0\)

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

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

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

\(\,\,\,\,\,\,\text{Axis of Symmetry: }\displaystyle x=4\)

\(\textbf{15)}\) Identify the x-intercepts of \(y=x^2-14x+54 \).

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There are no x-intercepts

See Related Pages\(\)

\(\bullet\text{ All Conic Section Notes}\)
\(\,\,\,\,\,\,\,\,\)

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

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

\(\bullet\text{ Axis of Symmetry}\)
\(\,\,\,\,\,\,\,\,x=-\frac{b}{2a}…\)

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

\(\bullet\text{ Area of Ellipses}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\pi a b…\)

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

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

\(\bullet\text{ Conic Sections- Parametric Equations}\)
\(\,\,\,\,\,\,\,\,x=h+r \cos{t}\)
\(\,\,\,\,\,\,\,\,y=k+r \sin{t}…\)

\(\bullet\text{ Degenerate Conics}\)
\(\,\,\,\,\,\,\,\,x^2−y^2=0…\)

\(\bullet\text{ Andymath Homepage}\)

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