Consider the Line L(t) = <2+t,5-4t>. Then L intersects: The x- axis at the point (3.25,0) when t = 5/4

The y- axis at the point (0, 13) when t = -2

The parabola y = x^2 at the points ___ and ___ when t = and

The parabola at the points and when and

Step-by-step explanation:

We have the following line written in parametric form :

with ∈ IR.

In order to find the intersection between and the parabola we know that '''' is the x-coordinate of the line and '''' is the y-coordinate of the line. Now, to solve this problem we need to find the values of '''' in which the intersection occurs. We can do this by replacing the components    '''' and '''' of in the equation of the parabola ⇒

= ( x component , y component ) = ( x , y ) ⇒

In the parabola : ⇒

Solving the equation we find that :

Using the quadratic formula with

, and

We find that the two possible values for t :

    and  

are and

This values and are the values of the parameter t where the line intersects the parabola so we can find the points by replacing the values of the parameter in the equation :

and

The final answer is

The parabola at the points and when and

the answer is b

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)