Please help with graphing a quadratic function. y=(x-1)^2 Got stuck at y=(x-1)(x-1) and p=1, q=1 Which doesn't work, because you can't fix a vertex with one point on a graph.
What is Alberbra?
I don't know what your p and q are upposed to be.
Define a new variable x' = x-1. Your equation becomes
y = x'^2
The vertex of your parabola is at x=1, y=0.
That is one point of the graph. Other points are easily plotted.
The parameter p is often used to designate the distance from the vertex to the focus. In tis case, p = 1/4, since the standard parabola formula is y = 4px^2 , with p being the focal distance.
The standard quadratic is
y=a(x-h)²+k
where (h,k) is the vertex, and x=h is the axis of symmetry.
The zeroes (p,q) are at -h±√(k/a).
If a>0, the quadratic is concave upwards, i.e. the vertex is at the minimum.
If a<0, the graph is concave downwards, or the vertex is at the maximum.
In the case of y=(x-1)²,
a=1, h=1, k=0
The zeroes are coincident because k=0, and coincide with the vertex. This also means that the graph is tangent to the x-axis.
I will leave it to you to figure out the concavity and the maximum/minimum.
Post if you need more information.
To graph the quadratic function y = (x - 1)^2, you can follow these steps:
Step 1: Find the vertex
The vertex form of a quadratic function is given by y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. In this case, h = 1 (from (x - 1)) and k = 0 (since there is no constant term in the equation). So, the vertex is located at (1, 0).
Step 2: Determine the axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex. In this case, since the x-coordinate of the vertex is 1, the axis of symmetry is a vertical line x = 1.
Step 3: Find the y-intercept
To find the y-intercept, substitute x = 0 into the equation. Therefore, y = (0 - 1)^2 = 1. So, the y-intercept is at (0, 1).
Step 4: Plot additional points
To better understand the shape of the graph, you can plot a few more points. Choose a few x-values greater and less than 1, substitute them into the equation, and calculate the corresponding y-values.
For example:
When x = -1, y = (-1 - 1)^2 = 4. So, the point (-1, 4) is on the graph.
When x = 2, y = (2 - 1)^2 = 1. So, the point (2, 1) is on the graph.
Step 5: Graph the parabola
Using the information gathered, plot the points (1, 0), (0, 1), (-1, 4), and (2, 1) on a coordinate plane. Then draw a smooth curve passing through these points.
The resulting graph should be a U-shaped parabola, symmetric around the axis of symmetry x = 1, and opening upward with the vertex at (1, 0).
Remember, practice makes perfect, so feel free to experiment and plot more points to strengthen your understanding of quadratic functions.