1.
SHOW Work please for understanding
1. Without graphing, determine the vertex of the parabola that is the graph of a given function,
state whether the parabola opens upward on downward. h(x)=4(x+1)^2-9
2.Vertex (-1,-2):Point (1,2) find the rule of a quadratic function whose graph has the given
vewrtex and passes through the given point.
3.f(x)=(x+2)^2 , graph the parabola and find it's vertex and axis of symmetry.
Solution
1.as the coefficient of the x^2 is positive hence the parabola opens downward .. and its minimum
value will be obtained when 4(x+1)^2 will be 0 and that value will be equal to -9. hence the
minimum value of the function = -9 obtained at x= -1 2. let the function be (ax+b)^2 + c so
putting x=-1 we should get its vertex , or a(-1)+b=0 a=b and c=-2 [because vertex is (-1,-2)]
f(x)= (ax+a)^2-2 it shall pass through 1,2 2= 4(a^2)-2 a=1 hence the function = x^2 + 2x -1 3.
vertex = (-2,0) AXIS OF SYMMETRY x=-2 graph, parabola opening upwards vertex = -2,0
passing through(1,9) , (-5,9) slope is 2x+4