Find x- and y- Intercepts....1

[mathdad]

Find the x- and y- intercepts.

y = 7x^3 + 3x^2 - 21x - 9

Solution:

Let x = 0

y = 7(0)^3 + 3(0)^2 - 21(0) - 9

y = -9

The graph crosses the y-axis at the point (0, -9).

Let y = 0

0 = 7x^3 + 3x^2 - 21x - 9

Factor by grouping.

7x^3 + 3x^2 = Group A

x^2(7x + 3)

-21x - 9 = Group B

-3(7x + 3)

x^2(7x + 3) -3(7x + 3)

(x^2 - 3)(7x + 3)

Set each factor to 0.

x^2 - 3

x^2 = 3

sqrt{x^2} = sqrt{3}

x = sqrt{3}

7x + 3 = 0

7x = - 3

x = -3/7

This means the graph crosses the x-axis at the points
(-3/7, 0) and (sqrt{3}, 0).

Answers:

y-intercept: -9

x-intercept: -3/7 & sqrt{3}

Is any of this correct?

I meant sqrt{3} not sqrt{x}.

 

[skeeter]

This means the graph crosses the x-axis at the points
(-3/7, 0) and (sqrt{x}, 0).

Answers:

y-intercept: -9

x-intercept: -3/7 & sqrt{x}

x = -3/7 works ... sqrt{x}?

solve the factor $x^2-3 = 0$ again for its two roots ...

 

[mathdad]

x = -3/7 works ... sqrt{x}?

solve the factor $x^2-3 = 0$ again for its two roots ...

I meant sqrt{3} not sqrt{x}.

- - - Updated - - -

x^2 - 3 = 0

x^2 = 3

sqrt{x^2} = sqrt{3}

x = sqrt{3}

 

[MarkFL]

I meant sqrt{3} not sqrt{x}.

- - - Updated - - -

x^2 - 3 = 0

x^2 = 3

At this point, you can say:

\(\displaystyle x=\pm\sqrt{3}\) :D

 

[skeeter]

$x^2-3=0 \implies (x-\sqrt{3})(x+\sqrt{3})=0$

or ...

$x^2=3$

$\sqrt{x^2} = \sqrt{3}$

$|x| = \sqrt{3} \implies x = \pm \sqrt{3}$

 

[mathdad]

Why do we always get two answers when taking the square root?

For example, the square root of 4 is -2 and 2. Why?

 

[skeeter]

Why do we always get two answers when taking the square root?

For example, the square root of 4 is -2 and 2. Why?

have a look at this article from NCTM ...http://www.nctm.org/Publications/Mathematics-Teacher/2015/Vol109/Issue2/The-Root-of-the-Problem/

 

[HallsofIvy]

Why do we always get two answers when taking the square root?

For example, the square root of 4 is -2 and 2. Why?

Because the square of 2 is 4 and the square of -2 is 4.

If [tex]a^2= y[/tex] then [tex](-a)^2= y[/tex] also.

 

[mathdad]

Ok. Interesting.