Answer:
(3,0) and (5,0)
Step-by-step explanation:
First let's write the generic model of a parabola:
[tex]y = ax^2 + bx + c[/tex]
Now, using the points (0, 15) and (4, -1), we have:
(0, 15):
[tex]15 = 0a + 0b + c[/tex]
[tex]c=15[/tex]
(4, -1):
[tex]-1 =16a+4b+15[/tex]
[tex]16a + 4b = -16[/tex]
[tex]4a + b = -4[/tex]
The x-coordinate of the vertex is given by the equation:
[tex]x_v = -b/2a[/tex]
[tex]-b/2a = 4[/tex]
[tex]b = -8a[/tex]
Now, using this value of b in the equation [tex]4a + b = -4[/tex], we have:
[tex]4a - 8a = -4[/tex]
[tex]a = 1[/tex]
[tex]b = -8[/tex]
The x-intercepts are where we have y = 0, so:
[tex]0 = x^2 -8x + 15[/tex]
Using Bhaskara's formula, we have:
[tex]\Delta = b^2 - 4ac = 64 -60 = 4[/tex]
[tex]x = -b \±\sqrt{\Delta}/2a[/tex]
[tex]x_1 = (8 + 2)/2 = 5[/tex]
[tex]x_2 = (8 - 2)/2 = 3[/tex]
So the x-intercepts are (3,0) and (5,0).
