Can someone help a mediocre year 10 student? (1 Viewer)

[lpodnano]

Okay its math. Signpost 5.1-5.3. I wrote all the questions I can't figure out when studying in the holidays.
Exercise 5:02 Question 13. (Its graphs and i cant be bothered drawing the graphs and scanning it so I'll just give the necessary information. Please provide necessary working out.
13. The Parabola in the diagram has its vertex at (-1,-8) and it passes through the point (1,4). The equation of the parabola has the form y=ax^2+bx+c

b. Use the axis of symmetry to show b=2a and the equation of the parabola is of the form y=ax^2+2ax-5

c. Substitute the coordinates of the vertex or the point (1,4) to find the value of a.

d. What is the equation of the Parabola?

Q 14. Using method of question 13 to find the equation of each of the following parabolas. (I'll just write one, then I'll get the idea after I see the working out)
b. It is concave down. It's vertex is (2,0). It passes through the point (3,-2) and it's y intercept is -8.

Okay next one
5:06 Question 11.a ( curves of the form y=ax^3+d )
Find its equation:
It's a positive curve that passes through points (0,10) and (2,20)

Okay. 5:08 ( general graphs )
10. The points A(-2,0), B(0,4) and C(4,0) form the vertices of an acute angled triangle.
a. Find the equation of the perpendicular bisectors of the sides AB, BC, and AC.
b. Find the point of intersection of the perpendicular bisectors of the sides AB and AC.
c. Show that the perpendicular bisector of side BC passes through the point of intersection found in b.

11. Points A(-1,2) B(3,2) C(-3,-2) form a triangle.
a. Find the equations of the medians.
b. Find the point of intersections of two of the medians.
c. Show that the third median passes through the point of intersection of the other two.


:mad1:As you can see I have a problem with graphs. I can do other parts of math just fine but graphs is just...

You don't have to give me working out to all of them. I would greatly appreciated just one working out. THANK YOU !!!
BTW: Tell me if I'm not giving enough information seeing as how its graphs and I might miss some.

 

[omniscience]

we cannot

 

[lpodnano]

we cannot

lol thanks...

 

[Lukybear]

Q13:
... U didnt give us a), but i suppose you just use the graph to work out that y intercept is 5.

b)
-b/2a = -1 (vertext)
-b=-2a
b=2a
sub into original equn

y=ax^2 + 2ax -5

C). sub (1,4) for into equn:

4=a +2a -5
3a=9
a=3

d) a=3
b=6

y=3x^2 + 6x -5


Althought i still cannot comprehend how you acquired the y intercept without graph, i.e. when x=0

 

[Lukybear]

14b).

-b/2a = 2
-b = 4a
y= ax^2 - 4ax - 8
Sub in (3,-2)
-2=9a-12a-8
-2=a
b=8

thus:
y=-2x^2 + 8x - 8

 

[Lukybear]

Good old memories! Im surprised you didnt hav trouble with 14 D.

 

[lpodnano]

Good old memories! Im surprised you didnt hav trouble with 14 D.

Thank you =)

 

[nerdsforever]

Okay next one
5:06 Question 11.a ( curves of the form y=ax^3+d )
Find its equation:
It's a positive curve that passes through points (0,10) and (2,20)

Look at the coordinate (0,10). THis means that the y intercept is 10. Hence, the d in your equation is 10.
y = ax^3 + 10. Substitute (2,20) in your equation.
8a = 10
a = 0.8
0.8x^3 + 10 = y

lol seems dodgy..

 

[lpodnano]

Okay next one
5:06 Question 11.a ( curves of the form y=ax^3+d )
Find its equation:
It's a positive curve that passes through points (0,10) and (2,20)

Look at the coordinate (0,10). THis means that the y intercept is 10. Hence, the d in your equation is 10.
y = ax^3 + 10. Substitute (2,20) in your equation.
8a = 10
a = 0.8
0.8x^3 + 10 = y

lol seems dodgy..

thanks. It's right except a = 5/4 I think you divided it the other way around.