Math Help - Graphs And Quadratics

7. The diagram shows the graphs of y = x2 – x, y = x + 2, y = 8 and y = -2x + 4.

Use the graphs to find the solutions to:

(a) x2 – x = 0

(b) x2 – x = x + 2

(c) x2 – x = 8

(d) x2 – x = -2x + 4

(e) x2 – x – 8 = 0

(f) x2 + x = 4

Im not sure how to read off this kind of graph. Help please. How exactly do I find soloutions from this graph? Please be detailed as I need to answer all these questions.

2. Hi there

for $x^2 - x = 0$ you need to find graphically where $y = x^2 - x$ cuts the x-axis. It seems to cut through the points $x = 0,1$

With the next question you have to consider where $y= x^2 - x$ and $y = x + 2$ intersect each other.

They do not intersectat a value that is clearly defined on your graph but you can conclude that the solution lies on $x \in (-1,0)\cap(2,3)$

To get exact values in can solve this kind of equations using a little algebra, solve for x

$x^2 - x = x + 2$

moving everything to one side

$x^2 - 2x -2 = 0$

then completing the square

Completing the square - Wikipedia, the free encyclopedia

$(x^2 - 2x +(\frac{-2}{2})^2 )-2-1= 0$

$(x - 1)^2 -3 = 0$

$x = \pm\sqrt{3}+1$

$x \approx -0.7,2.7$

3. nicely done, pickslides. As a matter of fact, we were doing completing the square in my algebra I class today, but it was more advanced. We had to complete the square, to get it from:

$ax^2+by^2+cx+dy+e=0$ by the way, this is the standard equation of a circle.

then, we completed the square twice, to make it:

$(x-h)^2+(y-k)^2=r^2$ this is the center-radius form, in a circle. h is the x value of the radius. k is the y value of the radius. r is the radius of the circle. x and y are any given point that the circle touches.

4. Hi.

e.) X^2 - x - 8 =0

f.) x^2 + x = 4

are the only two I am stuck on. Need help!

5. Originally Posted by Sailee316
Hi.

e.) X^2 - x - 8 =0

f.) x^2 + x = 4

are the only two I am stuck on. Need help!

Are we trying to sketch these functions or just solve for x?

If you want to sketch the functions then I would complete the square which turns the function into turning point form.

$x^2-x-8$

Take the middle terms coeffecient, half it, then square it. This gives you a new 3rd term.

$(x^2-x +(\frac{-1}{2})^2)-8-(\frac{-1}{2})^2)$

simplifying this

$(x^2-x +\frac{1}{4})-8-\frac{1}{4}$

factoring the expression inside the bracket gives

$(x-\frac{1}{2})^2-\frac{33}{4}$

This function is

$y =x^2$

moved to the right $\frac{1}{2}$ and down $\frac{33}{4}$

6. Originally Posted by pickslides
Are we trying to sketch these functions or just solve for x?

If you want to sketch the functions then I would complete the square which turns the function into turning point form.

$x^2-x-8$

Take the middle terms coeffecient, half it, then square it. This gives you a new 3rd term.

$(x^2-x +(\frac{-1}{2})^2)-8-(\frac{-1}{2})^2)$

simplifying this

$(x^2-x +\frac{1}{4})-8-\frac{1}{4}$

factoring the expression inside the bracket gives

$(x-\frac{1}{2})^2-\frac{33}{4}$

This function is

$y =x^2$

moved to the right $\frac{1}{2}$ and down $\frac{33}{4}$
UPDATE: I have spoken to my teacher and he says I have to extrapolate to find the interception points for the two questions. I think this means move x^2 - x down to -8 and read off where they intercept. If the line of x^2 -x passes through the line twice I must write down two sets of x and y coordinates. Still need help with these two. Thanks

7. Sailee! Do you have a handle on this thing yet? If not, let me know!

8. Originally Posted by Sailee316
7. The diagram shows the graphs of y = x2 – x, y = x + 2, y = 8 and y = -2x + 4.

Use the graphs to find the solutions to:

(a) x2 – x = 0

(b) x2 – x = x + 2

(c) x2 – x = 8

(d) x2 – x = -2x + 4

(e) x2 – x – 8 = 0

(f) x2 + x = 4

Im not sure how to read off this kind of graph. Help please. How exactly do I find soloutions from this graph? Please be detailed as I need to answer all these questions.
Duplicate questions: http://www.mathhelpforum.com/math-he...-graphs-2.html