Question

Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that is the minimum value of the graph. C) x = 2 and x= 3 because the graph crosses the x-axis at 2 and 3. D) x = -3 and x = -2 because the graph crosses the x-axis at -3 and -2.

Answer 1

Answer:

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~~~