Question

Nellie is analyzing a quadratic function f(x) and a linear function g(x). Will they intersect? f(x) g(x) graph of the function f of x equals one half times x squared, plus 2 x g(x) 1 5 2 10 3 15 (6 points) Yes, at positive x-coordinates Yes, at negative x-coordinates Yes, at negative and positive x-coordinates No, they will not intersect

Answer 1

Hi! :)

If we substitute 13 to f(x) and get a value for x, then the two equations will intersect.

So,

f(x) = x^2 + 6x +10
13 = x^2 + 6x+ 10
x^2 + 6x - 3 = 0

Solving the quadratic equations:

The roots of the equation is
x = 0.4641 and x = -6.4541

Therefore, the line will intersect with the quadratic function.

Answer 2

Hence, they will intersect at positive x-coordinates.

Explanation:

We are given two functions f(x) and g(x) as:

`f(x)=(1)/(2)x^2+2` and
`g(x)=5x`.

As the table of values for g are given as:

x g(x)

1 5

2 10

3 15

so on solving for g(x) using two point.

Let g(x)=y

We use the slope intercept form as:

y=mx+c

where m is the slope of line and c is the y-intercept of the line.

now for x=1 ,y=5

5=m+c

and for x=2, y=10

10=2m+c

on solving the above two equations using method of elimination we have:

m=5 and c=0

Hence, g(x)=5x

Now we can see by solving the equation f(x)=g(x) i.e. by solving a quadratic equation formed by equating f(x) and g(x) that the graph intersects at x=0.4174 and x=9.5826.

(since by solving

`f(x)=g(x)\\\\(1)/(2)x^2+2=5x\\\\x^2+4=10x\\\\x^2-10x+4=0` and hence solving this quadratic equation).

Hence, they will intersect at positive x-coordinates.