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 the vertex \((3,5)\), the equation becomes \(y=a(x - 3)^2+5\).

Step2: Find the value of \(a\)

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

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Step3: Write the final equation of the parabola

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:

• Vertex: Plot the point \((3,5)\).
• \(y\)-intercept: Plot the point \((0,1)\).
• Axis of symmetry: The axis of symmetry is \(x = 3\). We can find another point by using the axis of symmetry. If \((0,1)\) is a point on the parabola, then the point symmetric to \((0,1)\) about the line \(x = 3\) is \((6,1)\) (since the distance from \(x = 0\) to \(x = 3\) is \(3\) units, so we move \(3\) units to the right of \(x = 3\)).
• Since \(a=-\frac{4}{9}<0\), the parabola opens downwards.

Answer:

The equation of the parabola is \(y =-\frac{4}{9}(x - 3)^2+5\). Plot the vertex \((3,5)\), \(y\) - intercept \((0,1)\), its symmetric point \((6,1)\) and draw a parabola opening downwards.