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The graph of y+ax^2 + bx + c is given below, where a, b, and c are integers. Find a. (I saw an example of how to solve this one but I still don't get it) Thanks for the help :D


 Oct 24, 2019
edited by onetwothree  Oct 24, 2019
 #1
avatar+217 
0

hey this is an aops question

-1,-2 is the vertex, so you can make it into the vertex form

y = a(x+1)^2 - 2

then plug in 0,-1

-1 = a - 2

a = 3

 Oct 25, 2019
 #2
avatar+25275 
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The graph of \(y=ax^2 + bx + c\) is given below, where \(a\), \(b\), and \(c\) are integers. Find \(a\).

 

 

You get three Points from the picture above:

  • Point 1 ( Vertex ): \(P_1 (-1,-2)\)
  • Point 2 : \(P_2(0,-1)\)
  • Point 3 ( symmetric to Point 2) : \(P_3(-2,-1)\)

\(\begin{array}{|lrcll|} \hline & y &=& ax^2 + bx + c \\ \hline P_2(0,-1): & -1 &=& a*0+b*0 + c \\ & \mathbf{c} &=& \mathbf{-1} \\ \hline P_1(-1,-2): & -2 &=& a*(-1)^2+b*(-1) -1\\ & -2 &=& a -b -1 \\ & b &=& a + 1 \\ & \mathbf{b} &=& \mathbf{a+1} \\ \hline P_3(-2,-1): & -1 &=& a*(-2)^2+b*(-2) -1 \\ & -1 &=& 4a -2b -1 \\ & 4a-2b &=& 0 \quad | \quad :2 \\ & 2a-b &=& 0 \quad | \quad b=a+1 \\ & 2a-(a+1) &=& 0 \\ & 2a-a-1 &=& 0 \\ & a-1 &=& 0 \\ & \mathbf{a} &=& \mathbf{1} \quad | \quad b= a+1=1+1=2 \\ \hline \end{array} \)

 

\(y=x^2+2x-1\)

 

 

laugh

 Oct 25, 2019
edited by heureka  Oct 25, 2019

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