Math work pls help :)
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
If we draw a right triangle, it would have legs 6 and 2.
If you fit the vertex and the y-intercept into that form of the parabola, what you get is this:
and
and 16a=16. Solve that a to get that a = 1. Now you have your true parabola equation in vertex form, which is:
or just
Follow my work here to find the x-intercepts:
(x-4)(x-4)-1=y and
Simplifying by combining like terms gives you this parabola:
In order to find the x-intercepts, or the roots of the parabola as they are also known as, you have to set y equal to 0 (because the x-intercepts are found when y = 0, right?)
Factoring that gives you (x-5)(x-3)=0 and that x-5=0 and x-3=0. Solving each of those simple equations gives you x=5 and x=3 when y=0. So the roots, in the form you require them, are (5, 0) and (3, 0)
Follow my work here to find the x-intercepts:
(x-4)(x-4)-1=y and
Simplifying by combining like terms gives you this parabola:
In order to find the x-intercepts, or the roots of the parabola as they are also known as, you have to set y equal to 0 (because the x-intercepts are found when y = 0, right?)
Factoring that gives you (x-5)(x-3)=0 and that x-5=0 and x-3=0. Solving each of those simple equations gives you x=5 and x=3 when y=0. So the roots, in the form you require them, are (5, 0) and (3, 0)