(45,0), (0, 1.39), and (18, 3.2) Use those to find the equation of the quadratic. x=ax^2+bx+c
This is quite an interesting problem (at least to me). Each of the 3 points contains info on x and y: In (45,0), x=45 and y=0. Substitute those values into the quadratic y=ax^2+bx+c. You'll come up with an equation in a, b and c. Do the same thing for the other two points. You'll end up with 3 equations in a, b and c. Your job is to solve this system of linear equations for a, b and c. Then write out the original quadratic, substituting your values of a, b and c.
Whats The Original Equation Look Like?
The "original equation" is your x=ax^2+bx+c. all you need to change here is that initial x: You meant y=ax^2+bx+c.
Oh Okay Thank You And May I Ask If These Look Alright: 2025a+45b+c c=1.39 3.2=324a+18b+c
@LannaBoo24: The first expression you've typed in is not an equation. Can you fix this, so that you DO have an equation there?
Un Square It?
Notice that you don't have an ' = ' sign in the first expression you've typed in. That means you have an expression, not an equation. What you are trying to say in writing this equation is that "when x=45, y=0." Can you now fix the first expression (make it an equation)?
2025a+45b+c=0. Now you have 3 equations in the three unknowns, a, b and c. Please solve for a, b and c.
C=1.39 So I Can Just Put That Into The Other Two Equations Like This: 2025a+45b+1.39=0 324a+18b+1.39=3.2 Right?
Sure! That's one of the faster ways of solving this problem. Now you need only find the values of a and b.
Okkie Dokkie c: Well On The 2nd Equation That Has To Be Equal To 0 Also Or Is It Okay Like It Is?
Actually, your second equation is 0a + 0b +1c = 1.39, so your c=1.39 is correct.
Opps. I Meant In The 324a+18b+1.39=3.2 Sorry.
That you can find your own mistakes shows that you understand a great deal of what is happening here. :)
You're on the right track. I need to get off the Internet now, unfortunately. Good luck!!
Thank You c:
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