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 [tex]\frac{1}{2} x^2-5x+4[/tex].
[tex]\frac{1}{2}(0)^2-5(0)+4[/tex]
[tex]0-0+4[/tex]
[tex]0+4[/tex]
[tex]4[/tex]
So the first blank is 4 since h(0)=4.
h(2) means to replace x with 2 in [tex]\frac{1}{2} x^2-5x+4[/tex].
[tex]\frac{1}{2} (2)^2-5(2)+4[/tex]
[tex]\frac{1}{2} (4)-10+4[/tex]
[tex]2-10+4[/tex]
[tex]-8+4[/tex]
[tex]-4[/tex]
So the second blank is -4 since h(2)=-4.
h(4) means to replace x with 4 in [tex]\frac{1}{2} x^2-5x+4[/tex].
[tex]\frac{1}{2}(4)^2-5(4)+4[/tex]
[tex]\frac{1}{2}(16)-20+4[/tex]
[tex]8-20+4[/tex]
[tex]-12+4[/tex]
[tex]-8[/tex]
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.
