Question

I am have trouble with these x-intercepts. Please help me.​

Answer 1

Answer:

1. x=8, x=2
2. no solution
3. no solution

Step-by-step explanation:

For the equation ...

  y = a(x -h)² +k

you can find the x-intercepts by setting y=0 and solving for x.

  0 = a(x -h)² +k

  -k = a(x -h)² . . . . . . subtract k

  -k/a = (x -h)² . . . . . divide by a

  ±√(-k/a) = x -h . . . . take the square root

  h ± √(-k/a) = x . . . . add h . . . . this is the general solution

__

So, for each of your problems, fill in the corresponding numbers and do the arithmetic. If (-k/a) is a negative number, the square root gives imaginary values, so there is "no solution".

1.  x = 5 ± √9 = {5 -3, 5 +3} = {2, 8} . . . . the x-intercepts are 2 and 8

2.  x = -3 ± √(-2) . . . . . . no solution; the roots are complex

3.  x = 5 ± √(-8/4) . . . . . no solution; the roots are complex

Answer 2

Answer:

Step-by-step explanation:

These are all done the exact same way.  I'll do the first one in its entirety, and you can do the rest, following my example.

Finding x-intercepts means that you find the places in the polynomial where the graph of the function goes through the x-axis.  Here, the y-coordinates will be 0.  To find these x-intercepts, you have to set y equal to 0 and then factor.  First, though, we need to know exactly what the polynomial looks like in standard form.  The ones you have are all in vertex form.  We find the standard form by first expanding the binomial, like this:

$0=(x-5)(x-5)-9$

FOIL those out to get

$x^2-10x+25-9=0$

Combine like terms to get

$0=x^2-10x+16$

Now we have to factor that.  I'll use regular old factoring, although the quadratic formula will work also.

In our quadratic, a = 1, b = -10 and c = 16

The product of a * c = 16.  The factors of 16 are:

1, 16

2, 8

4, 4

Some combination of those factors will give us a -10, the b term.  2 and 8 will work, as long as they are both negative.  -2 + -8 = -10.  Fit them into the polynomial with the absolute value of the largest number named first:

$x^2-8x-2x+16=0$

Now we group them by 2's without ever changing their order:

$(x^2-8)-(2x+16)=0$

and then factor out the common thing in each set of parenthesis.  The common thing in the first set of parenthesis is an x; the common thing in the second set is a 2:

$x(x-8)-2(x-8)=0$

Now the common thing is (x - 8), so we factor that out and group together in a separate set of parenthesis what's left over:

$(x-8)(x-2)=0$

By the Zero Product Property, either x - 8 = 0 or x - 2 = 0.  Solving the first one for x:

x - 8 = 0 so x = 8

Solving the second one for x:

x - 2 = 0 so x = 2

The 2 solutions are x = 2 and x = 8, choices a and d.