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My IXL
Learning
Assessment
Analyt
Y. 11 Write a quadratic function from its vertex and another point YGV
A parabola opening up or down has vertex 
(-2,0) and passes through 
(-3,(1)/(4)). Write equation in vertex form.
Simplify any fractions.
Submit
Work it out
Not feeling ready yet? These can help:
4
Graph quadratic functions in vertex form

(83)
I.
Solve two-step linear

My IXL\newlineLearning\newlineAssessment\newlineAnalyt\newlineY. 1111 Write a quadratic function from its vertex and another point YGV\newlineA parabola opening up or down has vertex (2,0) (-2,0) and passes through (3,14) \left(-3, \frac{1}{4}\right) . Write equation in vertex form.\newlineSimplify any fractions.\newlineSubmit\newlineWork it out\newlineNot feeling ready yet? These can help:\newline44\newlineGraph quadratic functions in vertex form\newline(83) (83) \newlineI.\newlineSolve two-step linear

Full solution

Q. My IXL\newlineLearning\newlineAssessment\newlineAnalyt\newlineY. 1111 Write a quadratic function from its vertex and another point YGV\newlineA parabola opening up or down has vertex (2,0) (-2,0) and passes through (3,14) \left(-3, \frac{1}{4}\right) . Write equation in vertex form.\newlineSimplify any fractions.\newlineSubmit\newlineWork it out\newlineNot feeling ready yet? These can help:\newline44\newlineGraph quadratic functions in vertex form\newline(83) (83) \newlineI.\newlineSolve two-step linear
  1. Vertex form of parabola: Vertex form of a parabola: y=a(xh)2+ky = a(x - h)^2 + k. Given vertex (h,k)=(2,0)(h, k) = (-2, 0), so the equation becomes y=a(x+2)2y = a(x + 2)^2.
  2. Use point to find aa: Use the point (3,14)(-3, \frac{1}{4}) to find the value of aa. Substitute x=3x = -3 and y=14y = \frac{1}{4} into the equation: 14=a(3+2)2\frac{1}{4} = a(-3 + 2)^2.
  3. Simplify equation: Simplify the equation: 14=a(1)2\frac{1}{4} = a(-1)^2. 14=a(1)\frac{1}{4} = a(1).
  4. Solve for a: Solve for a: a=14a = \frac{1}{4}.

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