Question

Helpppppppppp pleaseeeeee

Answer 1

Explanation:

the general function of a parabola is

y = a(x - h)² + k

(h, k) is the vertex point (max. or min.).

"a" is a stretching factor, and indicates, if the parabola opens upwards (a is positive) or downwards (a is negative).

for f(x) = 4x² a = 4 (positive, parabola opens upward).

for f(x) = x² a = 1 (positive, parabola opens upward).

in both cases clearly h = 0 and k = 0, so the vertex is in both cases (0, 0) : the origin.

since the parabola opens upwards in both cases, the vertex is in both cases the minimum.

there is no maximum (or defined to be +infinity).

f(x) = 4x²

is stretched upwards compared to f(x) = x².

the functional values for every x are much larger for 4x² than for x². so, everything is stretched higher up, as (0, 0) stays at (0, 0). 4×0 is still 0.

Answer 2

f(x) = 4x² faces up, same as its parent function f(x) = x²

f(x) = 4x² has the same minimum as its parent function f(x) = x² (0)

f(x) = 4x² is vertically stretched by a factor of 4 relative to its parent function f(x) = x²

Explanation:

x² and 4x² are both equations of a parabola. the general equation is of the form y = ax² + bx + c where a, b and c are constants

1. Upward or downward facing?

In the function y = x² b and c are 0 and the a coefficient is 1

The a coefficient of 4x^2 is 4

In general if a > 0, the parabola will face upward i.e. it will open upward.

So both x² and 4x² will face up

2. Minimum or Maximum?
The graph of f(x) = x² has a minimum of f(x) = 0 at x =0. So 4x² will also have a minimum at x = 0

We can determine minimum by differentiating the function and setting the differential equal to 0 and solving for x.

y' = f(x)' = (x2)' = 2x and is 0 when x =0 and also y = 0
(4x²)' = 8x and is also 0 when x = 0, y = 0

So the minimum is 0 for both functions and is at (0, 0)

3. Stretched or compressed?
If you multiply a function f(x) by a positive integer > 1 you will be stretching the original function vertically.

To understand what that means, let's take the value of y at x = 1

For f(x) = x², y = 1² = 1

For f(x) = 4x², y = 4(1²) = 4

So the original y value has been multiplied 4 times in the transformed function. In other words a point (x, y) in x² has now (x, 4y) in 4x². This is called a vertical stretch because the function increases at 4 times the rate as before (it will be skinner than the original graph))

If we had multiplied by a constant a such that 0 < a < 1 then the transformed graph will be vertically compressed ( it will be fatter than the original graph)

Graphs of f(x) = x² and g(x) = 4x² are attached

Also graphed h(x) = 0.5x² so you can see the differences

Hope that helps.