Going to need a little help on this concept =)
@MrNood
@triciaal
@mathmale
You are given the vertex of this parabola, so start with the "vertex form of the equation of a parabola:"\[x-h=a(y-k)^2\] for a horiz. parabola. Know what to do next? Hint: what is the value of h? of k?
Would it be the ordered pair?
You are given the coordinates of the vertex of this parabola, as well as one other point on the parabola. This is all the info you need with which to find the coefficient a. Think about this...what are the coordinates of the vertex, and what are their names? Where will you insert info from the other given point, so that you can finally find "a"?
Well you would subtract
Even if you're unsure of what to do, take the given info and insert it into the "vertex form of the equation of a parabola" that I've given you. I need to see your involvement.
x−h=a(y−k)2 x+3=a(y+1)^2
Is this correct
@mathmale
Ugh please
@welshfella
Try solving this. You have replaced h, k, x and y with numeric values. That leaves only "a" as an unknown. Find a. Write out the "vertex equation" of this parabola, using your value of "a." Note that this equation MUST still contain x and y, but not h and k. Now let y=0 and solve for x. Does your equation correctly predict x?
Ugh I'm still so confused don't you have to compare the vertex to a point 1 unit away from the vertex?
@mathmale @mathmate
the equation in vertex form is x = a(y + 1)^2 - 3 you need to find the value of a ( ais the coefficient of x^2) plug in x = 4 and y = 0 4 = a *1^2 - 3 a = ?
OK?
any questions?
do you see why we use x = 4 and y = 0 to find the value of a?
So A=7
yes
Yes, what about the ordered pair do we ignore that
Because you have the given x and y values plugged in, in order to solve
the ordered pair (-3,-1) was the vertex of the graph and that enabled us to write the vertex form of the equation.
Okay, I see
Mathmale has given all the necessary info to solve the problem
Yes, Thank you so much =)
yw
@mathmate since I was working with you before, can I just bother you for a sec
Perhaps you'd like to post a new question.
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