Using the graph "Square root of 7,interval and the line where x is twice y "

Question

Given below is the graph of the function $ y = x^2 − 4x − 3$, Using the Graph (iii) write down the interval of values of $x$ for which the function is increasing in the interval $−6 < y < 0$.

(iv) find the roots of the equation $x^2 − 4x − 3 = 0$ and hence obtain the value of $\sqrt{7}$ to the nearest first decimal place.

(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.

Answer 1

OK starting to find the $\sqrt{7}$ using the graph we find that the roots of the graph are -0.6 and +4.6

I Think we should use the quadratic formula here to derive $\sqrt{7}$

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

so we have

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

From there?

Answer 2

For 3):

So, to figure out which interval the function is increasing on, you need to trace the function drawn with a pencil (or look with your eyes).

When you trace the function from left to right, when your hand goes up, the intervals below on the $x$-axis are the intervals where the function is increasing.

As you trace the function from left to right, when your hand moves down, the intervals where your hand goes down are the intervals where the function is decreasing.

It should be clear from the above explanation that the interval on the $x$-axis where the function is decreasing is $[-1,2]$ and the interval on the $x$-axis where it is increasing is $[2,5]$.

For 4):

You correctly found in your answer that the roots are $2 + \sqrt{7}$ and $2 - \sqrt{7}$ using the quadratic formula. Now, you also said by guessing that one root is 4.6 and one is -0.6

Since $2 + \sqrt{7}$ is positive (since the expression is adding two positive numbers), it must be the $4.6$ root, so $2 + \sqrt{7} = 4.6$ roughly. That means $\sqrt{7} = 2.6$ roughly.

For 5): It looks like it was answered in the comments.