please help with this graph question. i dont even know where to start

[agile echo]

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Once you've got your answer and you're all set, you can close this thread by typing +close, and then you can click the Thank buttons to show your Helper some appreciation. · Apr 8, 2026, 12:02 PM

[grizzled tree]

<@&1309522179368419349>

[grizzled tree]

<@&1309522179368419349>

[prime aurora]

k so first whenever you're given such a question, you must realize that the roots of $x^2 - 7/2x - 3 = 0$ are with the intersection of lets say some line $ax+b$ and the original quadratic $y = -x^2 + 3x + 6$, so since they intersect, $-x^2 + 3x + 6 = ax+b$.

[lofty grotto]

Annie Maqionde

[prime aurora]

I'm going to type out the general steps for solving such a problem; follow them and tell me if you're confused.

[grizzled tree]

ok

[prime aurora]

1. We match the coefficients of $x^2$ in both equations. Ofc, we can't change the second one,$-x^2 + 3x + 6 = ax+b$(not good practise to), so we'll change the first equation(multiply it by -1): $-x^2 +7x/2 + 3 = 0$.

[lofty grotto]

Annie Maqionde

[prime aurora]

Now also observe this. The required roots of the second equation MUST satisfy the first equation, so $-x^2 + x(3-a) + (6-b) = -x^2 + 7x/2 + 3 = 0$, right?

[lofty grotto]

Annie Maqionde

[grizzled tree]

how can u times it by -1 because the 0 wil be the same so your just changing one side of the equation

[prime aurora]

the other side is zero, and $0\cdot -1 = 0$?

[lofty grotto]

Annie Maqionde

[grizzled tree]

yes but in that logic u could just do this

x = 0

100000000x = 0

that doesnt work

0 x 100000000000 = 0

[prime aurora]

who said it doesn't work?

the roots are the same.

for our purposes, those equations are equivalent.

[grizzled tree]

ok it was a bad example but i still think that your not allowed to change only 1 side?

[prime aurora]

we're changing BOTH sides

i chose the second equation BECAUSE it has a zero on the other side

[grizzled tree]

yea i suppose thats true

[prime aurora]

got it?

[prime aurora]

now back to what I was saying, can you now find a and b?

hint: compare coefficients of x and the constant terms

[grizzled tree]

hmm ill try

is it

b = 3

a = -0.5

[prime aurora]

$b= 3, a = -\frac{1}{2}$

[lofty grotto]

Annie Maqionde

[prime aurora]

now, you have your line, i.e. $y = ax + b$, i.e. $y = \frac{-x}{2} + 3$

[lofty grotto]

Annie Maqionde

[prime aurora]

got it?

[grizzled tree]

sort of

idk where u get the y = ax + b from

[prime aurora]

equation of a straight line?

[grizzled tree]

or where u knew to rearrarange into this thing

[grizzled tree]

oh ok

[grizzled tree]

can u explain this please

@prime aurora

..?

ok ig not

<@&1309522179368419349>

[grizzled tree]

you know the roots is where it crosses the x axis

ur method is right but im not sure why ur talking abt the roots?

[prime aurora]

because that's what the question wants?

[prime aurora]

I equate it to zero since at the roots, it cuts the x-axis

[grizzled tree]

where do uget thus -a -b stuff

i dont get it

[prime aurora]

we already know that $-x^2 + 3x + 6 = ax +b$ right?

[lofty grotto]

Annie Maqionde

[grizzled tree]

yes

[prime aurora]

so $-x^2 + x(3-a) + (6-b) = 0$

[lofty grotto]

Annie Maqionde

[grizzled tree]

oooooooooooh

ok i get it now

thanks

i didnt realise u just minused that to make it equal 0

+close

[agile echo]

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