MHB Max or Min curve on a graph question

[ai93]

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

b) Sketch the graph of the function $$x^{2}+5x-6$$ 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 $$x=2$$ on the curve of $$y=x^{2}+5x-6$$

MY SOLUTION Right so far?

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

$$\therefore x=6$$ or $$-1$$

b)
$$y=x^{2}+5x-6$$

$$\d{y}{x}$$ = $$2x+5$$

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

Sub x into equation

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

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

and $$\d{y^{2}}{^{2}x}$$ = 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 $$-\frac{49}{4}$$

c) No clue!

 

[Evgeny.Makarov]

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

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

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

$$\therefore x=6$$ or $$-1$$

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

b)
$$y=x^{2}+5x-6$$

$$\d{y}{x}$$ = $$2x+5$$

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

Sub x into equation

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

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

and $$\d{y^{2}}{^{2}x}$$ = 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 $$-\frac{49}{4}$$

This is stated a little vague.

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

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.