# 5.1 Quadratic functions  (Page 7/15)

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Access these online resources for additional instruction and practice with quadratic equations.

## Key equations

 general form of a quadratic function $f\left(x\right)=a{x}^{2}+bx+c$ standard form of a quadratic function $f\left(x\right)=a{\left(x-h\right)}^{2}+k$

## Key concepts

• A polynomial function of degree two is called a quadratic function.
• The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down.
• The axis of symmetry is the vertical line passing through the vertex. The zeros, or $\text{\hspace{0.17em}}x\text{-}$ intercepts, are the points at which the parabola crosses the $\text{\hspace{0.17em}}x\text{-}$ axis. The $\text{\hspace{0.17em}}y\text{-}$ intercept is the point at which the parabola crosses the $\text{\hspace{0.17em}}y\text{-}$ axis. See [link] , [link] , and [link] .
• Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See [link] .
• The vertex can be found from an equation representing a quadratic function. See [link] .
• The domain of a quadratic function is all real numbers. The range varies with the function. See [link] .
• A quadratic function’s minimum or maximum value is given by the $\text{\hspace{0.17em}}y\text{-}$ value of the vertex.
• The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See [link] and [link] .
• The vertex and the intercepts can be identified and interpreted to solve real-world problems. See [link] .

## Verbal

Explain the advantage of writing a quadratic function in standard form.

When written in that form, the vertex can be easily identified.

How can the vertex of a parabola be used in solving real-world problems?

Explain why the condition of $\text{\hspace{0.17em}}a\ne 0\text{\hspace{0.17em}}$ is imposed in the definition of the quadratic function.

If $\text{\hspace{0.17em}}a=0\text{\hspace{0.17em}}$ then the function becomes a linear function.

What is another name for the standard form of a quadratic function?

What two algebraic methods can be used to find the horizontal intercepts of a quadratic function?

If possible, we can use factoring. Otherwise, we can use the quadratic formula.

## Algebraic

For the following exercises, rewrite the quadratic functions in standard form and give the vertex.

$f\left(x\right)={x}^{2}-12x+32$

$g\left(x\right)={x}^{2}+2x-3$

$f\left(x\right)={\left(x+1\right)}^{2}-2,\text{\hspace{0.17em}}$ Vertex $\text{\hspace{0.17em}}\left(-1,-4\right)$

$f\left(x\right)={x}^{2}-x$

$f\left(x\right)={x}^{2}+5x-2$

$f\left(x\right)={\left(x+\frac{5}{2}\right)}^{2}-\frac{33}{4},\text{\hspace{0.17em}}$ Vertex $\text{\hspace{0.17em}}\left(-\frac{5}{2},-\frac{33}{4}\right)$

$h\left(x\right)=2{x}^{2}+8x-10$

$k\left(x\right)=3{x}^{2}-6x-9$

$f\left(x\right)=3{\left(x-1\right)}^{2}-12,\text{\hspace{0.17em}}$ Vertex $\text{\hspace{0.17em}}\left(1,-12\right)$

$f\left(x\right)=2{x}^{2}-6x$

$f\left(x\right)=3{x}^{2}-5x-1$

$f\left(x\right)=3{\left(x-\frac{5}{6}\right)}^{2}-\frac{37}{12},\text{\hspace{0.17em}}$ Vertex $\text{\hspace{0.17em}}\left(\frac{5}{6},-\frac{37}{12}\right)$

For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry.

$y\left(x\right)=2{x}^{2}+10x+12$

$f\left(x\right)=2{x}^{2}-10x+4$

Minimum is $\text{\hspace{0.17em}}-\frac{17}{2}\text{\hspace{0.17em}}$ and occurs at $\text{\hspace{0.17em}}\frac{5}{2}.\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}x=\frac{5}{2}.$

$f\left(x\right)=-{x}^{2}+4x+3$

$f\left(x\right)=4{x}^{2}+x-1$

Minimum is $\text{\hspace{0.17em}}-\frac{17}{16}\text{\hspace{0.17em}}$ and occurs at $\text{\hspace{0.17em}}-\frac{1}{8}.\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}x=-\frac{1}{8}.$

$h\left(t\right)=-4{t}^{2}+6t-1$

$f\left(x\right)=\frac{1}{2}{x}^{2}+3x+1$

Minimum is $\text{\hspace{0.17em}}-\frac{7}{2}\text{\hspace{0.17em}}$ and occurs at $\text{\hspace{0.17em}}-3.\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}x=-3.$

$f\left(x\right)=-\frac{1}{3}{x}^{2}-2x+3$

For the following exercises, determine the domain and range of the quadratic function.

$f\left(x\right)={\left(x-3\right)}^{2}+2$

Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[2,\infty \right).$

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