Question

graph a parabola whose vertex is at (3,5) with y - intercept at y = 1.

Explanation:

Step1: Write the vertex form of the parabola

The vertex form of a parabola is \(y = a(x - h)^2 + k\), where \((h,k)\) is the vertex. Given \((h,k)=(3,5)\), the equation becomes \(y=a(x - 3)^2+5\).

Step2: Find the value of \(a\)

The \(y\)-intercept is at \(y = 1\) (when \(x = 0\)). Substitute \(x = 0\) and \(y=1\) into \(y=a(x - 3)^2+5\):

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Step3: Write the final equation

Substitute \(a =-\frac{4}{9}\) back into the vertex - form equation. The equation of the parabola is \(y=-\frac{4}{9}(x - 3)^2+5\).

To graph the parabola:

• Plot the vertex \((3,5)\).
• Since \(a=-\frac{4}{9}<0\), the parabola opens downwards.
• The \(y\)-intercept is \((0,1)\).
• We can also find the \(x\)-intercepts by setting \(y = 0\):

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So the \(x\)-intercepts are approximately \(x = 3 + 3.35=6.35\) and \(x=3 - 3.35=- 0.35\).

Answer:

The equation of the parabola is \(y =-\frac{4}{9}(x - 3)^2+5\). Plot the vertex \((3,5)\), \(y\)-intercept \((0,1)\) and \(x\)-intercepts (approximately \((-0.35,0)\) and \((6.35,0)\)) and draw a smooth curve through these points (opening downwards).