Question

I am extremely confused, can anyone help?​

Answer 1

Answer:

So I did most of you table for you:

First blank:4

Second blank:-4

Third blank:-8

Fourth blank: Left to you: Just plug in 6 into 1/2x^2-5x+4 (I will check)

Fifth blank: Left to you: Just plug in 8 into 1/2x^2-5x+4 (I will check)

The graph is D.

Step-by-step explanation:

So they have a table to fill in and they tell you in the first row what they want you to plug in

So the table is asking us to answer this:

What is h(0),h(2),h(4),h(6), and h(8).

h(0) means to replace x with 0 in $\frac{1}{2} x^2-5x+4$.

$\frac{1}{2}(0)^2-5(0)+4$

$0-0+4$

$0+4$

$4$

So the first blank is 4 since h(0)=4.

h(2) means to replace x with 2 in $\frac{1}{2} x^2-5x+4$.

$\frac{1}{2} (2)^2-5(2)+4$

$\frac{1}{2} (4)-10+4$

$2-10+4$

$-8+4$

$-4$

So the second blank is -4 since h(2)=-4.

h(4) means to replace x with 4 in $\frac{1}{2} x^2-5x+4$.

$\frac{1}{2}(4)^2-5(4)+4$

$\frac{1}{2}(16)-20+4$

$8-20+4$

$-12+4$

$-8$

So the third blank is -8 since h(4)=-8

Maybe you can try the last 2 blanks in the table part. That is try computing h(6) and h(8). I will check it for you.

Now the points I have so far are (0,4) from the h(0)=4, (2,-4) from the h(2)=-4, and (4,-8) from the h(4)=-8.

I'm looking for a graph that goes through these points (0,4) , (2,-4) , and (4,-8).

By the way the only graphs that are worth looking at is C and D because they are open up.  I know my curve for h(x)=1/2x^2-5x+4 should be a parabola open up because 1/2 is positive and 1/2 is the coefficient of x^2.

So Graph C has y-intercept (0,10) not (0,4) so Graph C is not right.

Graph D has y-intercept (0,4).  It also goes through (2,-4) and (4,-8).  

I don't know if you notice but the x-axis and y-axis are going up by two's in each graph.