Write the quadratic function in standard form (if necessary) and sketch its graph.
Identify the vertex
h(x)=x^2-9
looks pretty standard to me.
To play around with functions and their graphs, you can visit wolframalpha.com and just type in your function. For example,
http://www.wolframalpha.com/input/?i=x^2-9
To write the quadratic function in standard form, we need to expand the expression and rearrange it. The given quadratic function is h(x) = x^2 - 9.
Expanding this expression, we get h(x) = x*x - 9.
Now, rearranging the terms, we get h(x) = x^2 - 9. This is already in standard form, where the quadratic term (x^2) comes first, followed by the linear term (none in this case), and the constant term (-9) at the end.
To sketch the graph, we can consider the general shape of a quadratic function, which is a parabola. The coefficient in front of the x^2 term determines whether the parabola opens upward or downward. In this case, since the coefficient is positive (1), the parabola opens upward.
To find the vertex of the parabola, we can use the formula h = -b/(2a), where a and b are the coefficients of the quadratic and linear terms, respectively. In our case, since there is no linear term, b = 0, so the formula simply becomes h = -0/(2*1) = 0.
Therefore, the vertex of the parabola is at (0, -9), which is on the y-axis.
To sketch the graph, we can plot the vertex (0, -9). Since the coefficient of the x^2 term is 1, the parabola is symmetric about the vertex. We can then plot a few more points to get an idea of the parabola's shape. For example, substituting x = -3 and x = 3 into the function gives us h(-3) = -6 and h(3) = -6. These points can be plotted as well.
Once we have the vertex and a few additional points, we can draw a smooth curve that passes through these points while maintaining the general shape of a parabola. The resulting graph will be a U-shaped curve, opening upward, with its vertex at (0, -9).