The vertex of this parabola is at (3, 5). When the y-value is 6, the x-value is -1. What is the coefficient of the squared term in the parabola's equation?

Accepted Solution

Answer:1/16Step-by-step explanation:If you plot the vertex and the point, you'll find that the point lies above the vertex, thus, the parabola opens upwards, so the vertex form of this parabola will be:[tex]4p(y-k)=(x-h)^2[/tex]In order to solve for the leading coefficient, we first need to solve for p.  We will fill in the vertex form with all the info we have:[tex]4p(6-5)=(-1-3)^2[/tex]which simplifies to[tex]4p(1)=(-4)^2[/tex] and4p = 16 sop = 4Now we can rewrite the parabola using this value of p:[tex]4(4)(y-5)=(x-3)^2[/tex] and[tex]16(y-5)=(x-3)^2[/tex]We divide both sides by 16 to get[tex]y-5=\frac{1}{16}(x-3)^2[/tex] and to finish it off properly:[tex]y=\frac{1}{16}(x-3)^2+5[/tex]As you can see, the leading coefficient is 1/16