Respuesta :
Answer:
[tex]y=\frac{1}{30}x^2+\frac{1}{30}x-1[/tex]
Step-by-step explanation:
Given:
The x-intercepts of the parabola are (5,0) and (-6,0) and y-intercept (0, -1).
Now, the equation of the parabola with given x-intercepts is given as:
[tex]y=a(x-x_1)(x-x_2)[/tex]
Where, [tex]x_1\ and\ x_2\ \textrm{are the values of x in x intercepts.}[/tex]
'a' is a constant.
Now, plugging in [tex]x_1=-6,x_2=5[/tex], we get:
[tex]y=a(x-(-6))(x-5)\\\\y=a(x+6)(x-5)[/tex]
Now, we have the y-intercept as (0, -1). So, plug in this point in the above equation and solve for 'a'. This gives,
[tex]-1=a(0+6)(0-5)\\\\-1=a\times 6\times -5\\\\-1=-30a\\\\a=\frac{-1}{-30}\\\\a=\frac{1}{30}[/tex]
Therefore, the equation of the parabola is given as:
[tex]y=\frac{1}{30}(x+6)(x-5)\\\\y=\frac{1}{30}(x^2+x-30)\\\\y=\frac{x^2}{30}+\frac{x}{30}-\frac{30}{30}\\\\y=\frac{1}{30}x^2+\frac{1}{30}x-1[/tex]
Hence, the equation of the parabola is [tex]y=\frac{1}{30}x^2+\frac{1}{30}x-1[/tex].
