Question

Which functions have a y-intercept that is greater than the y-intercept of the function g(x) = |x + 3| + 4? Check three options.

Answer 1

Answer:

h(x) = –5 |x| + 10

k(x)=1/4(x-4)^2+4

m(x)=1/4 |x-8| +6

Step-by-step explanation:

The given function is:

g(x) = |x + 3| + 4

At y-intercept x=0,

g(0) = |0 + 3| + 4

g(0) = 3 + 4=7

The y-intercept of this function is 7.

We look for the functions with y-intercepts greater than 7.

$f(x)=-2(x-8)^2$

$f(0)=-2(0-8)^2$

$f(0)=-128$

h(x) = –5 |x| + 10

h(x) = –5 |0| + 10=10

$j(x)=-4(x+2)^2+8$

$j(0)=-4(0+2)^2+8=-8$

$k(x)=\frac{1}{4}(x-4)^2+4$

$k(0)=\frac{1}{4}(0-4)^2+4=8$

m(x)=1/4 |x-8| +6

m(0)=1/4 |0-8| +6=8

Answer 2

Answer with explanation:

The given function is

  g(x)=|x+3|+4

The meaning of Y intercept is the distance between origin and Point where the curve cuts Y axis.

In , g(x), put x=0

g(0)=|0+3|+4

   =3+4

   =7

So, Length of Y intercept =7 unit

2.

f(x)=-2(x-8)²

f(0)=-2×(0-8)²

     = -2 × 64

      = -128

Length of Y intercept =-128 unit

3.

h(x)=-5|x|+10

h(0)=-5 × |0| +10

       =10

Length of Y intercept =10 unit

4.

j(x)=-4(x+2)²+8

j(0)=-4×(0+2)²+8

     =-4 × 4+8

    = -16 +8

    = -8

Length of Y intercept =-8 unit

4.

$\rightarrow k(x)=\frac{1}{4} \times (x-4)^2+4\\\\\rightarrow k(0)=\frac{1}{4} \times (0-4)^2+4\\\\\rightarrow k(0)= 4+4\\\\=8$

Length of Y intercept =8 unit

5.

$\rightarrow m(x)=\frac{1}{4} \times |x-8|+6\\\\\rightarrow k(0)=\frac{1}{4} \times |0-8|+6\\\\\rightarrow k(0)= 2+6\\\\=8$

Length of Y intercept =8 unit

⇒ h(x),k(x) and m(x) has y intercept greater than y-intercept of the function g(x) = |x + 3| + 4.