Answer:
B) The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25.
Step-by-step explanation:
Given that the function is:
[tex]y=f(x)=-x^2+25[/tex]
It is quadratic equation, therefore we can observe that it is a parabola.
The coefficient of [tex]x^{2}[/tex] is negative, therefore it opens downwards.
That means it will have a maximum value.
This maximum value is at the vertex and where x = 0.
Also, vertex is at x intercept.
x intercept is the value of x where we have y = 0 and
y intercept is the value of y where we have x = 0
So, let us put y = 0 first.
[tex]0=-x^2+25\\ \Rightarrow x^{2} =25\\\Rightarrow x =5, -5[/tex]
Therefore x intercepts are at (5, 0) and (-5, 0)
Now, let us put x = 0,
[tex]y=0+25\\\Rightarrow y =25[/tex]
Therefore y intercept is at (0, 25).
It is the point of maximum value (which is 25) and vertex of parabola as explained above.
Kindly refer to the attached image for the graph of given function.
Therefore, correct option is:
B) The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25.
