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Elena L [17]
3 years ago
13

Use the graph to write a linear function that relates y to X.

Mathematics
2 answers:
Masteriza [31]3 years ago
6 0

Linear function that relates y to x.

y = x

coordinates of y = x

|  \ \  x  \ | -2 \ |   \ -1 |  \  \ 0 \ | \ \ 1 \  | \  \ 2  \ |  \  \ 3 \ \  \  | \\ --------------\\ | \ \ y \ \ | -2 \ |   \ -1 |  \  \ 0 \ | \ \ 1 \  | \  \ 2  \ |  \  \ 3 \ \  |

---------------------------------------------------------------------------------------------------------

A linear function is a straight line following equation: y = mx + b

<em>                                                               where m is slope, b is y-intercept</em>

frosja888 [35]3 years ago
6 0

⚘ <em>Refer</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>attachment</em><em> </em><em>for</em><em> </em><em>graph</em><em> </em><em>and</em><em> </em><em>below</em><em> </em><em>for</em><em> </em><em>some</em><em> </em><em>details</em><em>!</em><em>~</em>

<h3>Details ⤵️</h3>

  • x-intercept = 0
  • y-intercept = 0
  • slope = 1
  • Domain = x∈R

<em>It</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>linear</em><em> </em><em>function</em><em>.</em><em> </em><em>Since</em><em>,</em><em> </em><em>the</em><em> </em><em>line</em><em> </em><em>is</em><em> </em><em>straight</em><em> </em><em>when</em><em> </em><em>it's</em><em> </em><em>plotted</em><em> </em><em>and</em><em> </em><em>the</em><em> </em><em>linear</em><em> </em><em>function</em><em> </em><em>is</em><em> </em><em>(</em><em>y</em><em>=</em><em>x</em><em>)</em>

Some X and Y coordinates are:

\begin{gathered}\boxed{\begin{array}{c|c}\boxed{\sf x} &\boxed{\sf y} \\ \sf -2 &\sf -2 \\ \sf -1 &\sf -1\\ \sf 0 &\sf 0 \\ \sf 1 &\sf 1\\ \sf 2 &\sf 2 \end{array}}\end{gathered}

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Which is the graph of f(x) = -(x + 3)(x + 1)?
Bas_tet [7]

Answer:

See attachment

Step-by-step explanation:

A function is given to us and we need to tell which graph represents the given function. The function given to us is,

\tt: \implies f(x) = -( x + 3 )( x + 1 )

Let's find out at which points do the graph Intersects x axis / finding the roots. For that substitute f(x) = 0 , we have ,

\tt: \implies  -(x +3)( x + 1 ) = 0

Equate each factor by 0 ,

\tt: \implies \boxed{\blue{ \tt x = -1,-3 }}

Therefore the graph will intersect x axis at x is equal to -1 and x is equal to -3 .

On looking at the given graphs in the options the second graph intersects x axis at -1 and -3 .

<u>Hence</u><u> </u><u>the </u><u>second</u><u> </u><u>option</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u><u> </u>

{ See attachment }

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