What are the coordinates of the vertex for the graph of f(x) = 3x^2 + 6x - 9?

[Peter G.]

Hi

So: Let f (x) = 3x² + 6x - 9

For the graph of f:

a) Write down the coordinate of the vertex
b) Write down the equation of the axis of symmetry
c) Write down the the y intercept
iv) Find both x intercepts

My answers:
a) For a I managed to find the x coordinate of the vertex by:

x = -b / 2a
x = -6 / 6
x = -1

However, I am having difficulties with the y coordinates. This is what I tried:

c-b²/4a, hence: -9 - 36 = 45 / 12

But I am pretty sure I went wrong somewhere!

b) x = -1

c) The y intercept: - 9
d) The x's I factorized and got the answers, so my real problem is with the y coordinate of the vertex I guess!

Thanks,
Peter G.

 

[eumyang]

a) For a I managed to find the x coordinate of the vertex by:

x = -b / 2a
x = -6 / 6
x = -1

However, I am having difficulties with the y coordinates. This is what I tried:

c-b²/4a, hence: -9 - 36 = 45 / 12

Yep, it's wrong. The 4a part doesn't go under the c - b², just the b². It should be
$$c - \frac{b^2}{4a} = -9 - \frac{6^2}{4 \cdot 3} = -9 - \frac{36}{12} = ...$$

Anyway, why use this formula? I would just plug in the x-value into the original equation f(x) = 3x² + 6x - 9 and find the y-coordinate:
$$f(-1) = 3(-1)^2 + 6(-1) - 9 = ...$$

 

[Mark44]

Hi

So: Let f (x) = 3x² + 6x - 9

For the graph of f:

a) Write down the coordinate of the vertex
b) Write down the equation of the axis of symmetry
c) Write down the the y intercept
iv) Find both x intercepts

My answers:
a) For a I managed to find the x coordinate of the vertex by:

x = -b / 2a
x = -6 / 6
x = -1

However, I am having difficulties with the y coordinates. This is what I tried:

c-b²/4a, hence: -9 - 36 = 45 / 12

But I am pretty sure I went wrong somewhere!

After you have found the x-coordinate of the vertex, just evaluate the function at that number.

f(-1) = 3(-1)² + 6(-1) - 9 = 3 - 6 - 9 = -12.

Also, rather than memorize a formula for the coordinates that you will probably forget in time, it's probably better to find the vertex by completing the square.

y = 3x² + 6x - 9 = 3(x² + 2x) - 9 = 3(x² + 2x + 1) - 9 - 3 = 3(x + 1)² - 12

This shows that the vertex of the standard parabola y = x² has been shifted left by 1 unit, and shifted down by 12 units, which puts the vertex of the transformed function at (-1, -12).

b) x = -1

c) The y intercept: - 9
d) The x's I factorized and got the answers, so my real problem is with the y coordinate of the vertex I guess!

Thanks,
Peter G.

 

[Peter G.]

Thanks a lot guys!

Firstly, thanks for the formula I had wrong, and, I ended up realizing how stupid I was for not sub'ing the x value into the equation :shy:

 

[HallsofIvy]

Hi

So: Let f (x) = 3x² + 6x - 9

For the graph of f:

a) Write down the coordinate of the vertex
b) Write down the equation of the axis of symmetry
c) Write down the the y intercept
iv) Find both x intercepts

My answers:
a) For a I managed to find the x coordinate of the vertex by:

x = -b / 2a
x = -6 / 6
x = -1

However, I am having difficulties with the y coordinates. This is what I tried:

c-b²/4a, hence: -9 - 36 = 45 / 12

But I am pretty sure I went wrong somewhere!

It is better to think about what you are doing than just apply memorized formulas, especially here where you have mis-memorized the formula.
Since you know that x= -1, just use the original formula to find y= 3(-1)^2+ 6(-1)- 9

b) x = -1

c) The y intercept: - 9
d) The x's I factorized and got the answers, so my real problem is with the y coordinate of the vertex I guess!

Thanks,
Peter G.

Do you know where formulas you cite for the x and y coordinates of the vertex come from? Completing the square:
$$ax^2+ bx+ c= a(x^2+ (b/a)x )+ c= a(x^2+ (b/a)x+ (b^2/4a)- (b^2/4a))+ c$$
Taking that $-(b^2/4a)$ out of the parentheses, it is multiplied by that leading "a":
[tex]= a(x^2+ (b/a)x+ (b^2/4a)+ c- b^2/4= a(x+ b/2a)^2+ c- b^2/4[/itex]

When x= -b/2a, $y= (0)^2+ c- b^2/4a$. Since a square is never less than 0, for any other value of x, y is either larger than that (if a> 0) or less (if a< 0). In either case, x= -b/2a, $y= c- b^2/4$ are the x and y components of the vertex. (And notice that y is NOT $c-b^2/4a$

 

[Peter G.]

Hi, sorry, another doubt arose...

For: y = 3^x - 5

Find y when: x = 0, ± 1, ± 2x

So, for 0 and ± 1 I managed to do it, but I didn't understand the 2x

In addition to that, it then asks us to discuss y as x moves towards infinity and as x moves towards negative infinity. They expect me to say that, for positive infinity, it gets infinitely bigger but never smaller than -4 and for negative infinity I am unsure...

 

[Mark44]

If we define g(x) = 3^x - 5, we can see that:
g(0) = 3⁰ - 5 = 1 - 5 = -4
g(1) = 3¹ - 5 = 3 - 5 = -2
g(-1) = 3^-1 - 5 = 1/3 - 5 = -14/3

Then what is g(2x)? g(-2x)?

As x gets larger and larger, what does 3^x do?
As x gets more and more negative, what does 3^x do? To answer this, it might be helpful to calculate a few more values of g(x) for negative values of x.

 

[Peter G.]

Hi Mark 44, thank you for your patience

g(2x) would be for when x = 2 for example, 3⁴ - 5 = 76?

And, for when x is positive, as it increases, it keeps growing and growing.
As for when the values of negative x get bigger and bigger, 3^x keep getting smaller and smaller

 

[eumyang]

In addition to that, it then asks us to discuss y as x moves towards infinity and as x moves towards negative infinity. They expect me to say that, for positive infinity, it gets infinitely bigger but never smaller than -4 and for negative infinity I am unsure...

It sounds like you are being asked two things:
- as the x-values approaches positive infinity, what do the y-values approach?
- as the x-values approaches negative infinity, what do the y-values approach?

I think you already solve the first part (when x -> +∞, y -> +∞). The "never smaller than -4" has nothing to do with the first part, however. For the second part, try plugging in values for x (x = -1, x = -10, x = -100,...) and see what the corresponding y-values are.

EDIT: didn't see Mark44's post. Also:

As for when the values of negative x get bigger and bigger, 3^x keep getting smaller and smaller

You need to be more specific. Are the corresponding y-values getting larger and larger in the negative direction? Or they "leveling" off, getting close to a specific value?

 

[Mark44]

No, g(2x) is different from g(2), which is 3² - 5 = 9 - 5 = 4.
When you evaluate g(2x) you will get an expression that involves x.

As for when the values of negative x get bigger and bigger, 3^x keep getting smaller and smaller

Can you be more specific? "Smaller and smaller" could mean more and more negative.

 

[Peter G.]

Ok, for the 2x, I am still a bit confused. Would it be then: 3^2x - 5
= (3²)^x - 5
= 9^x - 5?

And, for what I meant with my statement: So, as the value of x gets increasingly smaller: e.g: -1, -2, -5, 3^x decreases in magnitude: 1/3, 1/9, 1/243. When 5 is subtracted from them, they start to level off, getting closer and closer to 5 but never quite reaching it?

 

[Mark44]

Ok, for the 2x, I am still a bit confused. Would it be then: 3^2x - 5
= (3²)^x - 5
= 9^x - 5?

Or you could just stop at 3^2x - 5

And, for what I meant with my statement: So, as the value of x gets increasingly smaller: e.g: -1, -2, -5, 3^x decreases in magnitude: 1/3, 1/9, 1/243. When 5 is subtracted from them, they start to level off, getting closer and closer to 5 but never quite reaching it?

A better way to say this is that as x gets more and more negative, 3^x gets closer to 0, but remaining positive. Subtracting 5 results in number that are getting closer to -5 (your earlier post had -4, which might have been a typo), but always staying above -5.

 

[Peter G.]

Ok, cool. Thanks!

Once again, thanks to all of you who contributed. You guys are great!