part ii only

y = mx + 1
y = 2x - x²

Putting these two equations together,

mx + 1 = 2x - x²

Note that putting them together does not mean that they necessarily equal each other; in fact, what we are actually trying to do is to put them on the same axes (which is the "putting them together" part) to see if they actually intersect.

x² + mx - 2x + 1 = 0
x² + (m - 2) x + 1 = 0

For them to meet at two distinct points,

Discriminant > 0
(m - 2)² - 4 (1) (1) > 0
m² - 2m - 2m + 4 - 4 > 0
m² - 4m > 0
m (m - 4) > 0

Solving this quadratic inequality gets us

m < 0 or m > 4

The values of m for which the line is a tangent to the curve are simply these boundary values, m = 0 and m = 4.

why is m<0 instead of >0 tho, since the entire equation is >0

Quadratic inequalities do not work in the same way as linear inequalities do.

In linear inequalities such as m - 4 > 0, we can easily solve this inequality as m > 4 because the graph of y = m - 4 lies entirely above the x-axis when m > 4, hence the solution m > 4.

In quadratic inequalities, the behaviour of the graph is different. The graph of y = m (m - 4) does not behave the same way as lines do. They actually start out from a high positive before going down to reach a minimum point and rebounding upwards back into a high positive value.

You will notice, based on this, that m (m - 4) is positive when m is to the left of the first m-intercept or to the right of the second m-intercept.

Try putting a value to confirm this.

First, we pick a value smaller than m = 0. Say, m = -5. Then, m (m - 4) = -5 (-5 - 4) = 45. This is a positive value, which is what we wanted.

Next, we pick a value between m = 0 and m = 4. Say, m = 2. Then, m (m - 4) = 2 (2 - 4) = -4. This is a negative value, so we disregard this.

Finally, we pick a value greater than m = 4. Say, m = 5. Then, m (m - 4) = 5 (5 - 4) = 5. This is a positive value, which is what we wanted.

Howsoever you try to input the values of m, you will realise that for the regions m < 0 or m > 4, the output value of m (m - 4) is positive, so these form our solutions to the inequality.

do all quadratic inequalities work the same way, left of first intercept and right of second intercept are positive

when solving quadratic inequalities do we have to put in values to see whether the arrows will be > or < ? as the arrows are where i usually go wrong

As long as the left-hand side of my quadratic inequality is in the form ax^2 + bx + c where a is positive and the right-hand side of the same inequality is zero, then all the time the positive regions are when

x < value of first intercept or
x > value of second intercept

while the solve negative region is the sandwiched region value

value of first intercept < x < value of second intercept,

with any equal sign being included as well where required.

Remember that we need to find the intercepts first before attempting to solve any equation or inequality at your level.