Question

30 POINTS PLEASE HELP!!! 4. The following equations represent the same quadratic function written in standard, vertex, and intercept form, respectively. f (x)=0.5x^2 +x-1.5, f (x)=0.5 (x+1)^2 -2, f (x) =(0.5x+1.5) (x-1) Based on these equations, which ofthe following is a trait of the graph of f (x) ? A: the range is y>= -2 B: the line of symmetry is x=0.5 C: the graph falls toward negative infinity to both the left and right D: the y-intercept is (0, -2) Answer correctly and I'll mark Brainliest! Thank you in advance, i was caught on this practice problem

Answer 1

Answer:

A

Step-by-step explanation:

So we have the quadratic equation and it's written in three equivalent forms:

$f(x)=0.5x^2+x-1.5\\f(x)=0.5(x+1)^2-2\\f(x)=(0.5x+1.5)(x-1)$

Let's determine the characteristics of the quadratic equation with the given equations.

From the first equation, since the leading coefficient (0.5) is positive, we can be certain that the graph opens upwards.

Also, the constant term is -1.5, so the y-intercept is (0,-1.5).

The second equation is the vertex form. Vertex form has the format:

$f(x)=a(x-h)^2-k$

Where (h,k) is the vertex. From the second equation we know that h is -1 (because (x+1) is the same as (x-(-1))) and k is -2. Therefore, the vertex is (-1,-2).

And since the graph points upwards, this means that (-1,-2) is the minimum point of the function. In other words, the range of the function is greater than or equal to -2. In interval notation, this is:

$[-2,\infty)$

This also means that the end behavior of the graph as a x approaches negative and positive infinity is positive infinity because the graph will always go straight up.

Also, the third form is the factored form. With that, we can solve for the zeros of the quadratic. The zeros are:

$0.5x+1.5=0\text{ and } x-1=0\\0.5x=-1.5 \text{ and }x=1\\x=-3\text{ and }x=1$

Therefore, the graph crosses the x-axis at x=-3 and x=1.

So, from the three equations, we gathered the following information:

1) The graph curves upwards.

2) The roots of zeros of the function is (-3,0) and (1,0).

3) The y-intercept is (0,-1.5).

4) The vertex is (-1,-2). This is also the minimum point.

5) Therefore, the range of the graph is all values greater than or equal to -2.

6) The end behavior of the graph on both directions go towards positive infinity.

Therefore, our correct answer is A.

B is not correct because the line of symmetry (or the x-coordinate of the vertex) here is -1 and not 1/2.

C is not correct because the graph goes towards positive infinity since it shoots straight up.

And D is not correct because the y-intercept is (0,-1.5).

Answer 2

Step-by-step explanation:

So we have the quadratic equation and it's written in three equivalent forms:

\begin{gathered}f(x)=0.5x^2+x-1.5\\f(x)=0.5(x+1)^2-2\\f(x)=(0.5x+1.5)(x-1)\end{gathered}

f(x)=0.5x

2

+x−1.5

f(x)=0.5(x+1)

2

−2

f(x)=(0.5x+1.5)(x−1)

Let's determine the characteristics of the quadratic equation with the given equations.

From the first equation, since the leading coefficient (0.5) is positive, we can be certain that the graph opens upwards.

Also, the constant term is -1.5, so the y-intercept is (0,-1.5).

The second equation is the vertex form. Vertex form has the format:

f(x)=a(x-h)^2-kf(x)=a(x−h)

2

−k

Where (h,k) is the vertex. From the second equation we know that h is -1 (because (x+1) is the same as (x-(-1))) and k is -2. Therefore, the vertex is (-1,-2).

And since the graph points upwards, this means that (-1,-2) is the minimum point of the function. In other words, the range of the function is greater than or equal to -2. In interval notation, this is: