Quadratics Help!

[TheSnark]

Hey guys. Im a new member and have a problem that i would really like help with. I can do quadratics if they're like y=2x^2-4x+1 but with the 1/2 im really confused.
Heres the equation.
I need the T.P., the x intercepts and the y int which i assume is -8
y=1/2(x+3)^2-8

Thanks in advance

 

[Deleted member 4993]

Hey guys. Im a new member and have a problem that i would really like help with. I can do quadratics if they're like y=2x^2-4x+1 but with the 1/2 im really confused.
Heres the equation.
I need the T.P., the x intercepts and the y int which i assume is -8
y=1/2(x+3)^2-8

Thanks in advance

If you expand the given quadratic we get:

y = 1/2 * x² + 3 * x - 7/2

Does that help.....

 

[Deleted member 4993]

> If you expand the given quadratic we get:
> y = 1/2 * x² + 3 * x - 7/2
How didya get that, Monsieur Lekhan?

Using PEMDAS....

 

[JeffM]

Hey guys. Im a new member and have a problem that i would really like help with. I can do quadratics if they're like y=2x^2-4x+1 but with the 1/2 im really confused.
Heres the equation.
I need the T.P., the x intercepts and the y int which i assume is -8
y=1/2(x+3)^2-8

Thanks in advance

For reasons that are no longer clear to me, students frequently find that fractions in algebra drive them bonkers. Until you are comfortable with fractions in algebra, I suggest getting rid of them as soon as possible and bringing them back if necessary only as the last step. So to get rid of that bothersome fraction, multiply your equation by 2. Now you have 2y = (x + 3)^2 - 16, which means 2y = x^2 + 6x + 9 - 16 = x^2 + 6x - 7. Easy, peasy. No fractions. However, as a last step, to eliminate the 2 from 2y you multiply the equation by (1/2) getting y = (1/2)(x^2 + 6x - 7), which is equivalent to Subhotosh Khan's answer.

 

[Deleted member 4993]

For reasons that are no longer clear to me, students frequently find that fractions in algebra drive them bonkers. Until you are comfortable with fractions in algebra,

I suggest getting rid of them as soon as possible ← You have been influenced by Denis too much


and bringing them back if necessary only as the last step. So to get rid of that bothersome fraction, multiply your equation by 2. Now you have 2y = (x + 3)^2 - 16, which means 2y = x^2 + 6x + 9 - 16 = x^2 + 6x - 7. Easy, peasy. No fractions. However, as a last step, to eliminate the 2 from 2y you multiply the equation by (1/2) getting y = (1/2)(x^2 + 6x - 7), which is equivalent to Subhotosh Khan's answer.

.

 

[JeffM]

.

Denis, who spends all his time in the corner with me, has become my mentor and gave me exclusive title to what used to be the Denis Rule. However, if someone does not find out why my cursor does not fully work at this site, I may start eliminating more things than just fractions: civil discourse may be one of them.

 

[TheSnark]

Got It!!!

Thanks very much guys for the help ive got it now. Ill definatly be using this forum again! Thanks again.

 

[Deleted member 4993]

y=1/2(x+3)^2-8
What I did was: 1/2(x+3)^2 - 8 = 0
(x + 3)^2 - 16 = 0 : x^2 + 6x + 9 - 16 = 0 : x^2 + 6x - 7 = 0 : (x - 1)(x + 7) = 0 : x = 1 or -7

What about:

need the T.P., the x intercepts and the y int which i assume is -8

 

[Mrspi]

Hey guys. Im a new member and have a problem that i would really like help with. I can do quadratics if they're like y=2x^2-4x+1 but with the 1/2 im really confused.
Heres the equation.
I need the T.P., the x intercepts and the y int which i assume is -8
y=1/2(x+3)^2-8

Thanks in advance

I'm going to take a different approach than the previous responders.

I'm betting you have been learning some different forms for the equation of a parabola.

ONE of those forms is commonly known as "vertex form," and, in general, looks like this:

y = a(x - h)² + k

where (h, k) is the vertex of the quadratic function so represented.

Ok...now you are asked for the "TP".....and I'm assuming that this stands for "turning point"????

The "turning point" of a parabola is at its vertex. So, for a parabola y = a(x - h)² + k, the turning point would be the point with coordinates (h, k).

For the x-intercept, you would substitute 0 for y, and solve the resulting equation for x.
For the y-intercept, you would substitute 0 for x, and solve the resulting equation for y.

Your equation is

y = (1/2)(x + 3)² - 8

compare that to

y = a(x - h)² + k

Can you identify the values of h and k in YOUR equation? What, then, would be (h, k), the turning point or vertex, for your parabola?

Then, to find the x-intercept(s), substitute 0 for y:

0 = (1/2)(x + 3)² - 8
8 = (1/2)(x + 3)²
Can you continue this to find the value(s) of x? (Hint...you should get two values for x)

To find the y-intercept, substitute 0 for x:
y = (1/2)(0 + 3)² - 8
Solving this for y is easy....it is just a matter of doing the arithmetic on the right side.

Just another (and I think simpler) approach to your problem.