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Write the equation of the parabola that passes through the points (2,24)(2,24), (4,0)(4,0), and (2,0)(-2,0). Write your answer in the form y=a(xp)(xq)y = a(x - p)(x - q), where aa, pp, and qq are integers, decimals, or simplified fractions.

Full solution

Q. Write the equation of the parabola that passes through the points (2,24)(2,24), (4,0)(4,0), and (2,0)(-2,0). Write your answer in the form y=a(xp)(xq)y = a(x - p)(x - q), where aa, pp, and qq are integers, decimals, or simplified fractions.
  1. Find pp and qq: We have the points (2,24)(2,24), (4,0)(4,0), and (2,0)(-2,0). The points (4,0)(4, 0) and (2,0)(-2, 0) are the xx-intercepts of the parabola, so they give us the values of pp and qq directly.\newlineqq00 and qq11.
  2. Write parabola equation: Now we can write the equation of the parabola in the form y=a(xp)(xq)y = a(x - p)(x - q) using the values of pp and qq we found.\newlineSubstitute 44 for pp and 2-2 for qq into the equation y=a(xp)(xq)y = a(x - p)(x - q).\newliney=a(x4)(x(2))y = a(x - 4)(x - (-2))\newliney=a(x4)(x+2)y = a(x - 4)(x + 2)
  3. Find value of aa: Next, we need to find the value of aa. We can use the point (2,24)(2,24) to do this, as this point lies on the parabola.\newlineSubstitute 22 for xx and 2424 for yy into the equation y=a(x4)(x+2)y = a(x - 4)(x + 2).\newline24=a(24)(2+2)24 = a(2 - 4)(2 + 2)
  4. Solve for aa: Now we solve the equation to find the value of aa.24=a(2)(4)24 = a(-2)(4)24=8a24 = -8aa=248a = \frac{24}{-8}a=3a = -3
  5. Final equation of parabola: We have found that a=3a = -3. Now we can write the final equation of the parabola using the values of aa, pp, and qq. Substitute 3-3 for aa, 44 for pp, and 2-2 for qq into the equation aa00. aa11

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