Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A parabola opening up or down has vertex (0,4)(0,4) and passes through (4,3)(4,3). Write its equation in vertex form.\newlineSimplify any fractions.\newline______

Full solution

Q. A parabola opening up or down has vertex (0,4)(0,4) and passes through (4,3)(4,3). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  1. Vertex Form of Parabola: What is the vertex form of the parabola?\newlineThe vertex form of a parabola is given by the equation y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Equation with Vertex: What is the equation of a parabola with a vertex at (0,4)(0, 4)?\newlineSubstitute 00 for hh and 44 for kk in the vertex form.\newliney=a(x0)2+4y = a(x - 0)^2 + 4\newliney=ax2+4y = ax^2 + 4
  3. Use Point to Find aa: Use the point (4,3)(4, 3) to find the value of aa. Replace the variables with (4,3)(4, 3) in the equation. Substitute 44 for xx and 33 for yy. 3=a(4)2+43 = a(4)^2 + 4 3=16a+43 = 16a + 4
  4. Solve for aa: Solve for aa.3=16a+43 = 16a + 4 Subtract 44 from both sides.34=16a3 - 4 = 16a1=16a-1 = 16a Divide both sides by 1616.116=a-\frac{1}{16} = a
  5. Equation with a=116a=-\frac{1}{16}: What is the equation of the parabola if a=116a = -\frac{1}{16}?\newlineSubstitute 116-\frac{1}{16} for aa in the equation y=ax2+4y = ax^2 + 4.\newliney=(116)x2+4y = (-\frac{1}{16})x^2 + 4\newlineVertex form of the parabola: y=116x2+4y = -\frac{1}{16}x^2 + 4

More problems from Write a quadratic function from its vertex and another point