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2 years ago

Help me plsssssssss its easy and show work please

Mathematics

1 answer:

babymother [125]2 years ago

8 0

Answer:

» For x-intercept, y is zero

 y = x² - 12x + 32 

${ \tt{0 = {x}^{2} - 12x + 32 }} \\ { \tt{ {x}^{2} - 12x + 32 = 0 }} \\ { \tt{(x - 8)(x - 4) = 0}} \\ { \tt{x = 8 \: \: and \: \: 4}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (8, \: 0) \: \: and \: \: (4, \: 0) \: \: }}}$

 y = x² - 10x +24 :

${ \tt{0 = {x}^{2} - 10x + 24}} \\ { \tt{ {x}^{2} - 10x + 24 = 0}} \\ { \tt{(x - 6)(x - 4) = 0}} \\ { \tt{x = 6 \: \: and \: \: 4}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (6, \: 0) \: \: and \: \: (4, \: 0) \: \: }}}$

 y = x² + 10x + 16 :

${ \tt{0 = {x}^{2} + 10x + 16}} \\ { \tt{ {x}^{2} + 10x + 16 = 0 }} \\ { \tt{(x + 2)(x + 8) = 0}} \\ { \tt{x = - 2 \: \: and \: \: - 8}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: ( - 2, \: 0) \: \: and \: \: ( - 8, \: 0) \: \: }}}$

 y = x² - 25 :

${ \tt{0 = {x}^{2} - 25}} \\ { \tt{ {x}^{2} = 25}} \\ { \tt{ \sqrt{ {x}^{2} } = \sqrt{25} }} \\ { \tt{x = \pm5}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (5, \: 0) \: \: and \: \: ( - 5, \: 0) \: \: }}}$

 y = x² + 8x + 12 :

${ \tt{0 = {x}^{2} + 8x + 12 }} \\ { \tt{ {x}^{2} + 8x + 12 = 0 }} \\{ \tt{(x + 2)(x + 6) = 0}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: ( - 2, \: 0) \: \: and \: \: ( - 6, \: 0) \: \: }}}$

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A) Graph the function and label all zeros. Be sure to label the axis and what you are incrementing by.

Liono4ka [1.6K]

9514 1404 393

Answer:

graph is shown below
absolute max and min do not exist
local max: 0 at x=0
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Step-by-step explanation:

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g(x) = x^2(x -5)

so has zeros at x=0 (multiplicity 2, meaning this is a local maximum*) and x=5.

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_____

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