Question

madieme96find the x-intercepts of the parabola with vertex (5,-4) and the y-intercept (0,96) and i have to write my answer in this form (x1,y1), (x2,y2)
and if necessary round to the nearest hundredth

Answer 1

Okay so the easiest way to solve this is by graphing it

Answer 2

hey another find the x-intercept question...
Find the x-intercept of the parabola with vertex (4,75) and y-intercept (0,27)
\[(x _{1},y _{1}), (x _{2},y _{2})\]

Answer 3

A parabola in vertex form is
\[y=a(x-h)^2+k,\text{ where the vertex is given by }(h,k).\]
For your first problem, you have
\[y=a(x-5)^2-4.\]
You're given the y-intercept is (0, 96). Plug in x = 0 and y = 96, and you'll be able to solve for \(a\):
\[96=a(0-5)^2-4\\
92=a(-5)^2\\
92=25a\\
a=\frac{92}{25}\]
So, the first parabola is \(y=\frac{92}{25}(x-5)^2-4.\)
The x-intercepts have coordinates of the form \((\overline{x},0).\) (The over line is just some notation to indicate any value of x.) To find this value of x, plug in y = 0 and solve for x:
\[0=\frac{92}{25}(x-5)^2-4\\
4=\frac{92}{25}(x-5)^2\\
\frac{100}{92}=(x-5)^2\\
\pm\sqrt{\frac{100}{92}}=x-5\\
x=5\pm\sqrt{\frac{100}{92}}\\
x=5\pm\frac{10}{2\sqrt{23}}\]
So, the x-intercepts of the first parabola are
\(\displaystyle\left(5+\frac{10}{2\sqrt{23}},0\right)\text{ and }\left(5-\frac{10}{2\sqrt{23}},0\right).\)

Do the same for the next question.