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

Answer

Explanation:

Step1: Recall vertex form of parabola

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

Step2: Use y - intercept to find ( a )

The y - intercept is at ( (0,1) ) (since when ( x = 0 ), ( y = 1 )). Substitute ( x = 0 ), ( y = 1 ) into the equation: [ \begin{align*} 1&=a(0 - 3)^2+5\ 1&=9a + 5\ 9a&=1 - 5\ 9a&=-4\ a&=-\frac{4}{9} \end{align*} ] So the equation of the parabola is ( y = -\frac{4}{9}(x - 3)^2 + 5 ).

Step3: Plot the vertex and other points

Vertex: ( (3,5) )
Y - intercept: ( (0,1) )
To find another point, we can use the axis of symmetry ( x = 3 ). The point symmetric to ( (0,1) ) about ( x = 3 ) is ( (6,1) ) (since the distance from ( x = 0 ) to ( x = 3 ) is 3, so we move 3 units to the right of ( x = 3 ) to get ( x = 6 )).
We can also find the value of ( y ) when ( x = 3\pm3 = 0,6 ) (already used) or ( x = 3\pm6= - 3,9 ). For ( x=-3 ): [ \begin{align*} y&=-\frac{4}{9}(-3 - 3)^2+5\ &=-\frac{4}{9}(-6)^2+5\ &=-\frac{4}{9}\times36 + 5\ &=-16 + 5\ &=-11 \end{align*} ] So the point is ( (-3,-11) ), and its symmetric point about ( x = 3 ) is ( (9,-11) ).
Now, plot the vertex ( (3,5) ), the y - intercept ( (0,1) ), the point ( (6,1) ), and other points like ( (-3,-11) ) and ( (9,-11) ) (if needed) and draw the parabola opening downward (since ( a=-\frac{4}{9}<0 )).

Answer:

The parabola has the equation ( y = -\frac{4}{9}(x - 3)^2 + 5 ) with vertex at ( (3,5) ), y - intercept at ( (0,1) ), and symmetric points like ( (6,1) ), ( (-3,-11) ), ( (9,-11) ) plotted to form the graph.