Min -7, Interval -6 to 0, & Find Roots of \sqrt{7}

[mathlearn]

Hi,
View attachment 5815

I drew the graph

1. Minumum value of the function = -7

can you help me to write the values for which the function is increasing interval -6<y<0

also can you help me to find the roots of the equation and obtain $$\sqrt{7}$$ to the nearest decimal place.

Help me to proceed

Many Thanks (Smile)

 

[phymat]

Can you post an attempt? Also, this should have been posted in the geometry sub-forum.

 

[mathlearn]

I .First I drew the graph ✔

II.The first question asks the minimum value which I found using the graph '-7' ✔

II. Interval

IV. (an attempt) When y=0 roots -0.6 and 4.7

y = x²- 4x -3

0 = x²- 4x -3

Can you help me to find the $$\sqrt{7}$$ to the nearest decimal place and the values of the function for which x is increasing the in interval -6<y<0

Many Thanks (Smile)

 

[Greg]

IV.

For $ax^2+bx+c=0$, the quadratic formula for $x$ is

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

so we have

$$x^2-4x-3=0\implies x=\dfrac{4\pm\sqrt{28}}{2}=2\pm\sqrt7$$

Referring to your graph, $2+\sqrt7\approx4.6$, so $\sqrt7\approx2.6$.

Can you now attempt III, using the above result for the positive root of $x^2-4x-3=0$?

 

[mathlearn]

Can you explain on $$\sqrt{7}+2$$ on the RHS of the equation . and How did this happen? Can you explain :)

Referring to your graph, 2+$$\sqrt{7}$$≈4.6 , so $$\sqrt{7}$$≈2.6.

Many Thanks :)

 

[Greg]

$$x=\dfrac{4\pm\sqrt{28}}{2}=\dfrac{4\pm\sqrt{4\cdot7}}{2}=\dfrac{4\pm2\sqrt7}{2}=2\pm\sqrt7$$

O.k?

 

[mathlearn]

I agree :) but what I'm uncertain is of how did you get 2 +$$\sqrt{7}$$ to be 4.6, What was referred in the graph to derive that?

Many Thanks :)

 

[Greg]

Look on the positive x-axis, where y = 0. :)

 

[mathlearn]

:) So to sum up the way of finding the $$\sqrt{7}$$using the graph.

1.Use the quadratic formula on the function
All of the rest of steps are clear and you have shown in a very detailed manner how the substituted quadratic formula is equal to $$\sqrt{7}+ 2$$.

*What do you call this method of equaling the "top right part of the quadratic formula to a simplified top right part "

and my other question is why did you pick the positive x-axis instead of the negative ?

Many Thanks :)

 

[Greg]

What do you call this method of equaling the "top right part of the quadratic formula to a simplified top right part "

Simplification, I guess.

...and my other question is why did you pick the positive x-axis instead of the negative ?

I could have picked either; the positive x-axis seemed easier and clearer at the time.

 

[mathlearn]

:) Now help me do V

(v) by drawing a suitable straight line, write down the coordinates of a point on the graph of which the x coordinate is twice the y coordinate.

Using Desmos I drew a graph https://www.desmos.com/calculator/sfxxgenmpz

Can anyone help me

_______________________________________________________________________________________________________Also help me in III

write down the interval of values of x for which the function is increasing in the interval −6 < y < 0.

Many Thanks :)

 

[Greg]

A hint for V: $y=\dfrac12x$

 

[mathlearn]

:) Many Thanks,

I drew a straight but i don't no whether it's correct

(v) by drawing a suitable straight line, write down the coordinates of a point on the graph of which the x coordinate is twice the y coordinate.

Can anyone comment

Many Thanks :)

 

[Greg]

V:

Use Desmos to graph $y=x^2-4x-3$ and $y=\dfrac12x$, then try to approximate the point of intersection of the two graphs.

Do you understand why the equation $y=\dfrac12x$ is useful? In fact, we can precisely determine the point of intersection of the two graphs by solving

$$x^2-4x-3=\dfrac12x$$

for $x$ and then finding $y$.

 

[mathlearn]

:)

V:

Use Desmos to graph $y=x^2-4x-3$ and $y=\dfrac12x$, then try to approximate the point of intersection of the two graphs.

Drew the line $$y=\dfrac12x$$ and updated the graph in desmos.

[graph]cprxgj0agr[/graph]

But I don't see an perfect point of intersection :)

V:

Do you understand why the equation $y=\dfrac12x$ is useful? In fact, we can precisely determine the point of intersection of the two graphs by solving

$$x^2-4x-3=\dfrac12x$$

for $x$ and then finding $y$.

$$x^2-4x-3=\dfrac12x$$

So after simplification it gets [multiply both sides by 2 to cancel the 2 in the denominator of the fraction on RHS]

$$2x^2-8x-6=x$$

So $$x=1$$ I guess

$$\therefore \displaystyle y=\dfrac12x$$
$$\displaystyle y=\dfrac12*1$$
$$\displaystyle y=\dfrac12$$

So There after , I don't seem to get it that much

Many Thanks :)

 

[Greg]

From $2x^2-8x-6=x$ we get $2x^2-9x-6=0$. Can you continue?

Looking at your hand-drawn graph, the point of intersection of $y=\dfrac12x$ and $x^2-4x-3$ is near $(x,y)=(-0.6,-0.3)$.

 

[mathlearn]

Yes absolutely :)

From $2x^2-8x-6=x$ we get $2x^2-9x-6=0$. Can you continue?

Looking at your hand-drawn graph, the point of intersection of $y=\dfrac12x$ and $x^2-4x-3$ is near $(x,y)=(-0.6,-0.3)$.

(v) by drawing a suitable straight line, write down the coordinates of a point on the graph of which the x coordinate is twice the y coordinate.

points of intersection (-0.6,-0.3); Correct ? ✔

Many Thanks :)

Now can you help me on

(iii) write down the interval of values of x for which the function is increasing in the interval −6 < y < 0.

 

[Greg]

That should be correct if you're expected to use the graph you drew. If you're expected to provide an exact value you need to solve the quadratic equation I gave in my previous post.

Very good.

 

[mathlearn]

Can you help me to find the interval.

(iii) write down the interval of values of x for which the function is increasing in the interval −6 < y < 0.

Many Thanks :)

 

[Greg]

Any thoughts on how to begin?

 

[mathlearn]

Any thoughts on how to begin?

I marked the range in orange

I think it should be 3<x<4.2, I am having trouble in writing the range

(iii) write down the interval of values of x for which the function is increasing in the interval −6 < y < 0.

Any Ideas ?

Many Thanks :)

 

[Greg]

Your orange line is correct. $x\in(3,4.6)$ or $3<x<4.6$, depending on which notation you are expected to use.

 

[mathlearn]

Many Thanks (Smile)