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Natali [406]
3 years ago
13

ASAP HELP MEeeeeeeeee

Mathematics
1 answer:
zzz [600]3 years ago
6 0
The correct answer is D.

A function cannot have an x-value that corresponds to more than one y-value.
In this table, the x-value '8' corresponds to two separate y-values; 8 and 5, making it invalid as a function.
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D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts.  You can graph the function and see when the graph crosses the x-axis or solve for the x-values.  I will solve it via factoring and so:

f(x)=x^2+5x+6

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6.  Now let's think about all the factors of 6 we have: 6×1 and 2×3.  Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5?  Actually we can say 6-1=5 and 2+3=5.  Let's try both.

First let's use 6 and -1 and so:

x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

f(x)=x+2\\\\0=x+2\\\\x=-2

Case 2:

f(x)=x+3\\\\0=x+3\\\\x=-3

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.


~~~Brainliest Appreciated~~~

8 0
3 years ago
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