what am I doing wrong? I'm not getting the correct answers.

[olemary]

My equation is to find the vertex, graph and find x intercepts of:
y= 0.3x^2+ 1.8x -2

I came up with x=1.6875 y=1.891 but when it comes to graphing this out, I have the wrong figures than the book does. Can you explain the solving of this to me?

Thanks,
Mary

 

[jwpaine]

Hi.

You want to get your quadratic in standard form for graphing
here are the steps for this problem.

y = 0.3x^2 + 1.8x - 2

(factor out 0.3)
y + 2 = 0.3(x^2+6x)

(add (1/2)b^2 into the enpty brackets)
y+2+0.3() = 0.3(x^2+6x+())
y + 2 + 0.3(9) = 0.3(x^2 + 6x + 9)

(simplify the left side, change the right side to squared form)
y + 4.7 = 0.3(x+3)^2

you are left with y = 0.3(x+3)^2 - 4.7

you have a non-rigid transformation 0.3(x)^2
you have a horizontal shift to the left (x+3)^2
and you have a virticle shirt, down x^2 - 4.7

Your vertex would be (-3,-4.7) remember, x = -b/(2a)

Question to think about:
How can you use the completing the square method to find your x-intercepts?

 

[stapel]

...find the vertex, graph and find x intercepts of y= 0.3x^2+ 1.8x -2

I came up with x=1.6875 y=1.891...

I will guess that this point is the vertex. But how did you arrive that this value? For instance, did you plug into the formula, h = -b/(2a) and k = f(h)?

Please reply showing your work and reasoning. Thank you.

Eliz.

 

[soroban]

Re: what am I doing wrong? I'm not getting the correct answ

Hello, Mary!

Find the vertex and x-intercepts, and graph of:
. . \(\displaystyle y\:=\:0.3x^2\,+\,1.8x\,-\,2\)

Are you familiar with the "vertex formula"?

For the parabola: \(\displaystyle \,y\:=\:ax^2\,+\,bx\,+\,c\), the vertex is at: \(\displaystyle \,x\:=\:\frac{-b}{2a}\)


We have: \(\displaystyle \,a\,=\,0.3,\:b\,=\,1.8,\:c\,=\,-2\)

The vertex is at: \(\displaystyle \,x\:=\:\frac{-1.8}{2(0.3)} \:=\:-3\)

. . Then: \(\displaystyle \:y\:=\:0.3(-3)^2\,+\,1.8(-3)\,-\,2\:=\:-4.7\)

Therefore: the vertex is \(\displaystyle (-3,\,-4.7)\)


For the x-intercepts, set the function equal to zero and solve for \(\displaystyle x.\)

We have: \(\displaystyle \:0.3x^2\,+\,1.8x\,-\,2\:=\:0\)

Multiply 10: \(\displaystyle \:3x^2\,+\,18x\,-\,20\:=\:0\)

Quadratic Formula: \(\displaystyle \:x \:=\:\frac{-18\,\pm\,\sqrt{18^2\,-\,4(3)(-20)}}{2(3)} \:=\:\frac{-9\,\pm\,\sqrt{141}}{3}\)

The x-intercepts are: \(\displaystyle \:\left(\frac{-9\,-\,\sqrt{141}}{3},\,0\right)\,\) and \(\displaystyle \,\left(\frac{-9\,+\,\sqrt{141}}{3},\,0\right)\)

. . These are very close to \(\displaystyle (-7,0)\) and \(\displaystyle (1,0).\)

 

[stapel]

Please reply showing your work and reasoning.

Never mind that now....

Eliz.

 

[Denis]

Nice way to demonstrate, Soroban; I think olemary needed that.

Say thanks to Soroban, olemary :wink:

 

[olemary]

what am I doing wrong?

Nice way to demonstrate, Soroban; I think olemary needed that.

Say thanks to Soroban, olemary :wink:

I did! I even printed it out and am using it as my example in my textbook.
Thanks,
Mary
:?