Respuesta :
Given: A parabola that has a vertical axis, vertex (4,-6), and x-intercept of -3.
Required: To find the standard equation of a parabola.
Explanation: The equation of the parabola with vertex (h,k) and opening up (a>0) or down (a<0) is
[tex]y=a(x-h)^2+k[/tex]Putting (h,k)=(4,-6) we get
[tex]y=a(x-4)^2+(-6)[/tex]Now, the x-intercept is the point where the curve intersects the x-axis.
Hence, P(x,y)=(-3,0). Putting this in the equation of the parabola and solving for 'a' we get,
[tex]\begin{gathered} 0=a(-3-4)^2-6 \\ a=\frac{6}{49} \end{gathered}[/tex]Hence the equation of the parabola is
[tex]y=\frac{6}{49}(x-4)^2-6[/tex]This is the vertex form of the parabola. The standard form of the parabola is
[tex]\begin{gathered} y=\frac{6}{49}x^2-\frac{48}{49}x-\frac{198}{49} \\ \end{gathered}[/tex]Final Answer: The standard form of the parabola is
[tex]y=\frac{6}{49}x^2-\frac{48}{49}x-\frac{198}{49}[/tex]