Math Is Fun Forum

bingbong

Member

Registered: 2016-03-04

Posts: 8

Algebra- Graphing Quadratics

1. The grid lines in the graph below are one unit apart. The red parabola shown is the graph of the equation y = ax^2 + bx + c. Find a+b+c.
http://s24.postimg.org/iex99fmv9/prob_2.png

2. Find the vertex of the graph of the equation x -y^2 + 6y =8.

3. Find the area of the region enclosed by the graph of the equation x^2 + y^2 = 4x + 6y+13.

4. The graph of the equation y =ax^2 + bx + c, where a, b, and c are constants, is a parabola with axis of symmetry x = -3. Find b/a.

5. The grid lines in the graph are one unit apart. The red parabola shown is the graph of the equation y=ax^2 + bx + c. Find a*b*c.
http://s9.postimg.org/4q0l1oyi7/prob_7.png

6. The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equation x^2 + Bx + y^2 + Dy + F = 0 for some constants B, D, and F. Find B+D+F.

7. The point (a,b) is 5 units away from the point (6,3), and (a,b) lies on the line 5x - 4y = -14. What is the largest possible value of a?

I know it's a lot of problems so I'll be patient and wait as long as needed for help! Thx in advance guys!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra- Graphing Quadratics

Hi;

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,145

Re: Algebra- Graphing Quadratics

hi bingbong

I'm happy to help you through a method for each of these.  But let's take it one at a time.

Q1.  From the diagram we can see this graph goes through (2,1), (0,5) and (4,5)

So one way to do this would be to start with y = ax^2 + bx + c, substitute the three sets of values, and then solve as simultaneous equations.

But here's a neater way:

If the x axis was 5 higher then the points where the graph crosses the x axis would be (0,0) and (4,0)

So an equation for this would be y = (x-0)(x-4)  But it could also be stretched by a factor 'a' so let's start with y = ax(x-4)

Now shift it up 5 units so it goes through (0,5) and (4,5)  ... y = ax(x-4) + 5

So substitute in (2,1) to get 'a' and you're done.

If you post me your equation I'll then look at Q2.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bingbong

Member

Registered: 2016-03-04

Posts: 8

Re: Algebra- Graphing Quadratics

Thanks, I figured out how to solve some of the problems.
For Problem one, my equation was: y= a(x-2)^2 + 1

I only need help on Q4,5,and 6

Bob

Administrator

Registered: 2010-06-20

Posts: 10,145

Re: Algebra- Graphing Quadratics

hi bingbong

4. The graph of the equation y =ax^2 + bx + c, where a, b, and c are constants, is a parabola with axis of symmetry x = -3. Find b/a.

All quadratics have a line of symmetry which tells us the lowest (or highest) point the graph reaches.

From the quadratic formula the two roots are either side of -b/2a so you can put this equal to -3.

5. The grid lines in the graph are one unit apart. The red parabola shown is the graph of the equation y=ax^2 + bx + c. Find a*b*c.

This is like question 1.  You can pick out three points that the graph goes through.  You need to start with an equation like this:

y - -a(x-p)(x-q) where p and q are the x coordinates equally spaced either side of the line of symmetry.  Adjust to 'lift' the graph up to go through those two points and use the third to find 'a'.

Q6.  This is a circle centre (3,5) with radius 4.  So make a sketch and put in the line of reflection.  Work out the new centre (radius the same) and construct the new equation.

In general for any circle centre (a,b) and radius c

Please post back how you are doing so I can see if you are 'on track'.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob