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# Vertex Form of a Quadratic Equation

## Convert to vertex form of a quadratic equation by completing the square

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Vertex Form of a Quadratic Equation

### Vertex Form of a Quadratic Equation

Probably one of the best applications of the method of completing the square is using it to rewrite a quadratic function in vertex form. The vertex form of a quadratic function is

yk=a(xh)2\begin{align*}y - k = a(x - h)^2\end{align*}

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h,k)\begin{align*}(h, k)\end{align*}.

It is also simple to find the x\begin{align*}x-\end{align*}intercepts from the vertex form: just set y=0\begin{align*}y = 0\end{align*} and take the square root of both sides of the resulting equation.

To find the y\begin{align*}y-\end{align*}intercept, set x=0\begin{align*}x = 0\end{align*} and simplify.

#### Finding the Vertex and Intercepts of Parabolas

Find the vertex, the x\begin{align*}x-\end{align*}intercepts and the y\begin{align*}y-\end{align*}intercept of the following parabolas:

a) y2=(x1)2\begin{align*}y - 2 = (x - 1)^2\end{align*}

Vertex: (1, 2)

To find the x\begin{align*}x-\end{align*}intercepts,

Set y=0:Take the square root of both sides:22=(x1)2=x1and2=x1\begin{align*}\text{Set} \ y = 0: & & -2 & = (x - 1)^2 \\ \text{Take the square root of both sides}: & & \sqrt{-2} & = x - 1 && \text{and} && -\sqrt{-2} = x - 1\end{align*}

The solutions are not real so there are no \begin{align*}x-\end{align*}intercepts.

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y - 2 & = (-1)^2\\ \text{Simplify}: & & y - 2 & = 1 \Rightarrow \underline{y = 3}\end{align*}

b) \begin{align*}y + 8 = 2(x - 3)^2\end{align*}

\begin{align*}& \text{Rewrite}: & & y - (-8) = 2(x - 3)^2\\ & \text{Vertex}: & & \underline{(3, -8)}\end{align*}

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}\text{Set} \ y = 0: & & 8 & = 2 (x - 3)^2\\ \text{Divide both sides by} \ 2: & & 4 & = (x - 3)^2 \\ \text{Take the square root of both sides}: & & 2 & = x - 3 && \text{and} && -2 = x - 3\\ \text{Simplify}: & & & \underline{\underline{x = 5}} && \text{and} && \underline{\underline{x = 1}}\end{align*}

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y + 8 & = 2(-3)^2\\ \text{Simplify}: & & y + 8 & = 18 \Rightarrow \underline{\underline{y = 10}}\end{align*}

To graph a parabola, we only need to know the following information:

• the vertex
• the \begin{align*}x-\end{align*}intercepts
• the \begin{align*}y-\end{align*}intercept
• whether the parabola turns up or down (remember that it turns up if \begin{align*}a > 0\end{align*} and down if \begin{align*}a < 0\end{align*})

#### Graphing Parabolas

1. Graph the parabola given by the function \begin{align*}y + 1 = (x +3)^2\end{align*}.

\begin{align*}& \text{Rewrite}: & & y - (-1) = (x - (-3))^2\\ & \text{Vertex}: & & \underline{(-3, -1)} && \text{vertex}:(-3, -1)\end{align*}

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}&\text{Set} \ y = 0: & & 1 = ( x + 3)^2\\ &\text{Take the square root of both sides}: & & 1 = x + 3 \qquad \text{and} \qquad -1 = x + 3\\ &\text{Simplify}: && \underline{\underline{x = -2}} \qquad \quad \text{and} \qquad \quad \ \underline{\underline{x = -4}}\\ &&& x-\text{intercepts}: \ (-2, 0) \ \text{and} \ (-4, 0)\end{align*}

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}& \text{Set} \ x = 0: & & y + 1 = (3)^2\\ & \text{Simplify:} & & \underline{\underline{y = 8}} && y-\text{intercept}: (0, 8)\end{align*}

And since \begin{align*}a > 0\end{align*}, the parabola turns up.

Graph all the points and connect them with a smooth curve:

2. Graph the parabola given by the function \begin{align*}y = - \frac{1}{2} (x - 2)^2\end{align*}.

\begin{align*}& \text{Rewrite} & & y - (0) = - \frac{1} {2} (x - 2)^2\\ & \text{Vertex:} & & \underline{(2, 0)} && \text{vertex:} (2, 0)\end{align*}

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}\text{Set} \ y = 0: & & 0 & = - \frac{1} {2} (x - 2)^2 \\ \text{Multiply both sides by} \ -2: & & 0 & = (x - 2)^2 \\ \text{Take the square root of both sides}: & & 0 & = x - 2\\ \text{Simplify}: & & & \underline{\underline{x = 2}} && x-\text{intercept:} (2, 0)\end{align*}

Note: there is only one \begin{align*}x-\end{align*}intercept, indicating that the vertex is located at this point, (2, 0).

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y & = -\frac{1} {2}(-2)^2 \\ \text{Simplify:} & & y & = - \frac{1} {2} (4) \Rightarrow \underline{\underline{y = -2}} && y- \text{intercept:}(0, -2)\end{align*}

Since \begin{align*}a < 0\end{align*}, the parabola turns down.

Graph all the points and connect them with a smooth curve:

### Example

#### Example 1

Graph the parabola given by the function \begin{align*}y = 4(x +2)^2-1\end{align*}.

\begin{align*}& \text{Rewrite} & & y - (-1) = 4(x +2)^2\\ & \text{Simplify} & & y +1 = 4(x +2)^2\\ & \text{Vertex:} & & \underline{(-2, -1)} && \text{vertex:} (-2, -1)\end{align*}

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}\text{Set.} \ y = 0: & & 0 & = 4(x +2)^2-1 \\ \text{Subtract 1 from each side}: & & 1 & = 4(x +2)^2 \\ \text{Divide both sides by 4}: & & \frac{1}{4} & = (x +2)^2 \\ \text{Take the square root of both sides}: & & \frac{1}{2} & = \pm (x + 2)\\ \text{Separate}: & & & \frac{1}{2}=-(x+2) && \frac{1}{2}=x+2)\\ \text{Simplify}: & & & \underline{\underline{x = -2.5}} && \underline{\underline{x = -1.5}}\end{align*}

The \begin{align*}x-\end{align*}intercepts are \begin{align*}(-2.5, 0)\end{align*} and \begin{align*}(-1.5, 0)\end{align*}.

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y & = 4(0 +2)^2-1 \\ \text{Simplify:} & & y & = 15 \Rightarrow \underline{\underline{y = 15}} && y- \text{intercept:}(0, 15)\end{align*}

Since \begin{align*}a < 0\end{align*}, the parabola turns up.

Graph all the points and connect them with a smooth curve:

### Review

Rewrite each quadratic function in vertex form.

1. \begin{align*} y= x^2 - 6x\end{align*}
2. \begin{align*}y + 1 = -2x^2 -x\end{align*}
3. \begin{align*}y = 9x^2 + 3x - 10\end{align*}
4. \begin{align*}y = -32x^2 + 60x + 10\end{align*}

For each parabola, find the vertex; the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts; and if it turns up or down. Then graph the parabola.

1. \begin{align*}y - 4 = x^2 + 8x\end{align*}
2. \begin{align*}y = -4x^2 + 20x - 24\end{align*}
3. \begin{align*}y = 3x^2 + 15x\end{align*}
4. \begin{align*}y + 6 = -x^2 + x\end{align*}
5. \begin{align*}x^2-10x+25=y+9\end{align*}
6. \begin{align*}x^2+18x+81=y+1\end{align*}
7. \begin{align*}4x^2-12x+9=y+16\end{align*}
8. \begin{align*}x^2+14x+49=y+3\end{align*}
9. \begin{align*}4x^2-20x+25=y+9\end{align*}
10. \begin{align*}x^2+8x+16=y+25\end{align*}

RE

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### Vocabulary Language: English

TermDefinition
Intercept The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
Parabola A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
Vertex A vertex is a corner of a three-dimensional object. It is the point where three or more faces meet.