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Identify the vertex first. Then use it to find the axis of symmetry.
Graph:
Description: See solution.
We want to draw the graph of the given quadratic function. Note that the function is already written in vertex form y=a(x-h)^2+k, where a, h, and k are either positive or negative numbers. y=(x-2)^2-1 To draw the graph we will follow four steps.
We will first identify the constants a, h, and k. Recall that if a<0, the parabola will open downwards. Conversely, if a>0, the parabola will open upwards. Vertex Form:& y= a(x- h)^2+k Function:& y= 1(x- 2)^2+(- 1) We can see that a= 1, h= 2, and k=- 1. Since a is greater than 0, the parabola will open upwards.
Let's now plot the vertex ( h,k) and draw the axis of symmetry x= h. Since we already know the values of h and k, we know that the vertex is ( 2,- 1). Therefore, the axis of symmetry is the vertical line x= 2.
Note that both points have the same y-coordinate.
Finally, we will sketch the parabola which passes through the three points. Remember not to use a straightedge for this!
The domain of quadratic functions is all real numbers. We can see above that the minimum point of the curve is reached at the vertex. Thus, the range is all real numbers greater than or equal to - 1. Domain:& All real numbers Range:& y ≥ - 1 By looking at the graph we can state the values for the x- and y-intercepts. We see that the parabola intercepts the y-axis at y=3, and intercepts the x-axis at x=1 and x=3. $y-$intercept:& y=3 $x-$intercepts:& x=1 and x=3 We know that the shape of quadratic function is a parabola, and in the given function it opens upwards. Thus, the function decreases on the interval (- ∞, 2), it reaches the vertex at (2,- 1), and then it increases on the interval (2,∞ ).