Question

Can anyone please help with the attached file? I know the y-intercept is (0,-7) & I think the vertex is (-3/4,-65/8) but I'm not too sure on that one and I can't find the last part at all

Answer 1

@KeithAfasCalcLover do you think you could help me out? :)

Answer 2

So we know that to find the y-intercept, set x to zero:
\[f(x)=2x^2+3x-7\rightarrow f(0)=-7\]

Answer 3

yep, i got that one :)

Answer 4

To find the vertex we need to complete the square:
\[\eqalign{
f(x)&=2x^2+3x-7 \\
&=2(x^2+1.5x)-7 \\
&=2(x^2+1.5x+0.75^2-0.75^2)-7 \\
&=2(x^2+1.5x+0.75^2)-2(0.75^2)-7 \\
&=2(x+0.75)^2-2\left(\frac{3}{4}\right)^2-7 \\
&=2(x+0.75)^2-2\left(\frac{9}{16}\right)-\frac{112}{16} \\
&=2(x+0.75)^2-\frac{18}{16}-\frac{112}{16} \\
&=2(x+0.75)^2-\frac{130}{16} \\
&=2\left(x+\frac{3}{4}\right)^2-\frac{65}{8}
}\]
So therefore, the vertex is at:
\[V=\left(-\frac{3}{4},-\frac{65}{8}\right)\]

Answer 5

Good so far?

Answer 6

yes sir :) now i just need help with the last part " Use the quadratic formula to find the coordinates of the x-intercepts of the function's graph. Round your answers to the nearest hundredth. Express your answer in the form"

Answer 7

Haha: I like that "sir"; I can get used to that ;)

ANYWAYS lol, "To find the x-intercepts, set y to zero" So we have:
\[0=2x^2+3x-7\]
We can use the quadratic formula like so:
\[\eqalign{
&a=+2 \\
&b=+3 \\
&c=-7 \\
}
\phantom{..........}x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
So then:
\[x=\frac{-3\pm\sqrt{3^2-4(2)(-7)}}{2(2)}=\frac{-3\pm\sqrt{9+56}}{4}=\frac{-3\pm\sqrt{65}}{4}\]

Answer 8

You can approximate the two x-values lets call them \(x_A\) and \(x_B\):
\[x_A=\frac{{-3}+\sqrt{65}}{4}\approx1.27\phantom{space}x_B=\frac{{-3}-\sqrt{65}}{4}\approx-2.77\]
So then the co-ordinates would be:
\(x_A(1.27,0)\) and \(x_B(-2.77,0)\)

Answer 9

OMG thank you so much @KeithAfasCalcLover !! you are awesome!!! thank you for taking up a lot of your time to explain it so well :) do you think you could check another problem i had similar to this one? i have the answers but this is an important assignment so i want to make sure its correct :)

Answer 10

no he cant

Answer 11

Yeah I can of course! I was just having supper :)

Answer 12

And thanks Ali anytime! I would recommend though that you put it in another question and link me like this:
@Keithafascalclover
So that way we can close this one!