Max or Min curve on a graph question

[ai93]

a) Find the roots of the equation \(\displaystyle x^{2}+5x-6\)

b) Sketch the graph of the function \(\displaystyle x^{2}+5x-6\) labeling the points at which the graph crosses the axes and the co-ordinates of the maximum and minimum of the curve

c) Find the equation of the tangent at the point where \(\displaystyle x=2\) on the curve of \(\displaystyle y=x^{2}+5x-6\)

MY SOLUTION Right so far?

a) Using the quadratic formula, we get \(\displaystyle x=\frac{-5\pm\sqrt{49}}{2}\)

\(\displaystyle \therefore x=6\) or \(\displaystyle -1\)

b)
\(\displaystyle y=x^{2}+5x-6\)

\(\displaystyle \d{y}{x}\) = \(\displaystyle 2x+5\)

\(\displaystyle x=-\frac{5}{2}\) (-2.5)

Sub x into equation

\(\displaystyle y=(-\frac{5}{2})^{2}+5(-\frac{5}{2})-6\)

y=\(\displaystyle -\frac{49}{4}\) (-12.25)

and \(\displaystyle \d{y^{2}}{^{2}x}\) = 2 which is a minimum value

So with the graph, you would plot it with the parabola going with the x points -3 and 6 and the y points \(\displaystyle -\frac{49}{4}\)

c) No clue!

 

[Evgeny.Makarov]

a) Find the roots of the equation \(\displaystyle x^{2}+5x-6\)

Note that this is not an equation because an equation must have the = sign. You could find the roots of the equation \(\displaystyle x^{2}+5x-6=0\) or of polynomial \(\displaystyle x^{2}+5x-6\).

a) Using the quadratic formula, we get \(\displaystyle x=\frac{-5\pm\sqrt{49}}{2}\)

\(\displaystyle \therefore x=6\) or \(\displaystyle -1\)

I recommend substituting these values back into the original equation and checking if they are indeed roots.

b)
\(\displaystyle y=x^{2}+5x-6\)

\(\displaystyle \d{y}{x}\) = \(\displaystyle 2x+5\)

\(\displaystyle x=-\frac{5}{2}\) (-2.5)

Sub x into equation

\(\displaystyle y=(-\frac{5}{2})^{2}+5(-\frac{5}{2})-6\)

y=\(\displaystyle -\frac{49}{4}\) (-12.25)

and \(\displaystyle \d{y^{2}}{^{2}x}\) = 2 which is a minimum value

I agree. It may make sense to remember that the $x$-coordinate of the vertex of the parabola $y=ax^2+bx+c$ is $-b/(2a)$ and that the type of extremum is determined by the sign of $a$: if $a>0$, then the function has a minimum and if $a<0$, then it has a maximum.

So with the graph, you would plot it with the parabola going with the x points -3 and 6 and the y points \(\displaystyle -\frac{49}{4}\)

This is stated a little vague.

c) Find the equation of the tangent at the point where \(\displaystyle x=2\) on the curve of \(\displaystyle y=x^{2}+5x-6\)

The equation of the tangent to $y=f(x)$ at point $(x_0,f(x_0))$ is $y-y_0=f'(x_0)(x-x_0)$.

 

[ai93]

The equation of the tangent to $y=f(x)$ at point $(x_0,f(x_0))$ is $y-y_0=f'(x_0)(x-x_0)$.

Is this the formula? What values would you need to sub in?

 

[Evgeny.Makarov]

Is this the formula?

Well, it's not an apple! (Smile) You can see it in Wikipedia.

What values would you need to sub in?

I think my post mentions everything that one needs to use this formula. I suggest you re-read it, and if you have further questions, feel free to describe what you don't understand.

 

[CellCoree]

c) To find the equation of the tangent at the point where x=2, we first need to find the corresponding y-value on the curve. Plugging in x=2 into the original equation, we get y=2^{2}+5(2)-6=8. So the point on the curve is (2,8).

Next, we need to find the slope of the tangent at this point. We can do this by finding the derivative of the original equation, which we already did in part b). The derivative is \d{y}{x}=2x+5. Plugging in x=2, we get a slope of 9.

Now we can use the point-slope form of a line to find the equation of the tangent. The point-slope form is y-y_{1}=m(x-x_{1}), where m is the slope and (x_{1},y_{1}) is the point on the line. Plugging in our values, we get y-8=9(x-2). Simplifying, we get y=9x-10 as the equation of the tangent at the point where x=2.