I really need help on this question please!

Mathematics

1 answer:

Ymorist [56]3 years ago

7 0

Okay, I know this looks intimidating but don't worry, it isn't that hard! All you need to do is take any two points on the line, we'll do (-3, 0) and (0,6), choose one to be x1 and y1, and then the other to be x2 and y2, it doesn't matter which one you decide to choose. We'll use (-3,0) as x1 and y1 then (0,6) as x2 and y2, now just use the formula, which is y2 - y1 / x2 - x1, to use this take y2(6) and y1(0) and subtract 6 - 0, which equals six. Then we'll use x2(0) - x1(-3) which equals 3, so now, you take y2 - y1 /x2 - x1 and turn it into 6/3, and you get two, hope this helped you!

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2. Find the stationary points for the function

vovikov84 [41]

Answer & step-by-step explanation:

Stationary points are the points where the first derivative is equal to zero.

Let's calculate it using the power rule (exponent comes forward, decrease exponent by 1) and the fact that the derivative is a linear operation (that is $D[a\ f(x) + b\ g(x)] = a Df(x) + b Dg(x)$ )

The first derivative is then

$y' = \frac13 (3x^2) - \frac12 (2x) -6 = x^2-x-6 = (x+2)(x-3)$

Note that the last passage is not strictly needed, but it's really helpful to find stationary points, when in this next passage we set it equal to zero. Alternatively, you can use the quadratic formula if you can pull the factors out of your head right away.

$y'=0 \rightarrow (x+2)(x-3) = 0 \implies x=-2 || x=3$

These two point could be maxima, minima, or inflection points. To check them you can either see how the sign of the first derivative goes, or check the sign of the second derivative, as you're required.

The rules states that if the second derivative evaluated in that point is negative we have a maximum, if it's positive we have a minimum, and if we have a zero we keep derivating until we get a non-zero value.

In our case, the second derivative we get by calculating the derivative again and we get $y'' = 2x-1$ . Evaluating it at both points we get