First, let's expand the given equation:
[tex]\text{ f(x) = }-(x-7)^2\text{ - }29[/tex][tex]\text{ f(x) = }-(x^2-14x+49)^{}\text{ - }29[/tex][tex]\text{ f(x) = }-x^2+14x-49\text{ - }29[/tex][tex]\text{ f(x) = }-x^2+14x-78[/tex]
We get, a = -1, b = 14 and c= -78
A.) Let's determine the vertex.
[tex]\text{ x = }\frac{-b}{2a}\text{ = }\frac{-(14)}{2(-1)}\text{ = }\frac{-14}{-2}[/tex][tex]\text{ x = 7}[/tex]
Substituting x = 7 in the equation, let's find y.
[tex]y\text{ = }-(x-7)^2\text{ - }29[/tex][tex]y\text{ = }-(7-7)^2\text{ - }29[/tex][tex]y\text{ = }-(0)^2\text{ - }29[/tex][tex]\text{ y = -29}[/tex]
Therefore, the coordinate of the vertex is x,y = 7, -29
B. Let's determine the equation of the axis of symmetry.
The equation for the axis of symmetry of a parabola can be expressed as:
[tex]\text{ x = }\frac{-b}{2a}[/tex]
We get,
[tex]\text{ x = }\frac{-b}{2a}\text{ = }\frac{-(14)}{2(-1)}\text{ = }\frac{-14}{-2}[/tex][tex]x\text{ = 7}[/tex]
Therefore, the equation for the axis of symmetry is x = 7.
