Quadratic function in vertex form:
[tex]y=a(x-h)^2+k[/tex]
Vertex: (h,k)
Given function:
[tex]y=-4(x-5)^2+125[/tex]Vertex: (5,125)
Axis of symmetry (x-value in the vertex): x=5
y-intercept: find y when x is 0:
[tex]\begin{gathered} y=-4(0-5)^2+125 \\ y=-4(-5)^2+125 \\ y=-4(25)+125 \\ y=-100+125 \\ y=25 \end{gathered}[/tex]
y-intercept: (0,25)
x-intercepts (solutions), equal the function to zero and solve x.
[tex]\begin{gathered} -4(x-5)^2+125=0 \\ -4(x^2-10x+25)+125=0 \\ -4x^2+40x-100+125=0 \\ -4x^2+40x+25=0 \\ \\ ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-40\pm\sqrt{40^2-4(-4)(25)}}{2(-4)} \\ \\ x=\frac{-40\pm\sqrt{1600+400}}{-8} \\ \\ x=\frac{-40\pm\sqrt{2000}}{-8} \\ \\ x_1=\frac{-40+\sqrt{2000}}{-8}\approx-0.6 \\ \\ x_2=\frac{-40-\sqrt{2000}}{-8}\approx10.6 \end{gathered}[/tex]
x-intercepts: (-0.6, 0) and (10.6, 0)
Graph: In green the axis of symmetry
Domain of a funtion is the set of x-values for which the function is defined
Range of a function is the set of values the function takes
Domain: All real numbers (-∞,∞)Range: From -infionite to 125 (-∞,125]In the given situation the vertex states the maximum height of the rocket (125) and the time it takes to reach that maximum (5 seconds)
The solutionsrepresents the moment (time) the rocket is in the ground; as it starts at a height of 25 (y-intercept) the negative solution doesn't have sence in the situation;
the solution (10.6, 0) means that after 10.6 seconds of the launch of the rocket it hits the ground
