Help find the vertex point of each quadratic equation?

Question

Please, please can someone help me find the vertex point, the graph's direction and any axis intercepts.

1. y=-x+1
2. y=x^2-4x
3. y=x+1

Answer 1

Equations #1 and #3 are straight lines, so they have no vertices. They do have x and y-intercepts however. The slope-intercept form for the equation of a line is y = mx + b. Look at the coefficient, m, of the x term. It will be the slope of the line. If m is positive, the line will rise from left to right. If m is negative, it will fall from left to right. The constant following the x term is the y-intercept, because if we let x = 0, then the graph of the line crosses the y-axis at the point where y = b. To find the x-intercept, let y = 0, and then solve for x. That value for x will be the point where the graph of the line crosses the x-axis.

#1)
y = -x + 1

m = -1, so the line falls from left to right.

To find y-intercept, let x = 0.
y = 0 + 1
y = 1

To find x-intercept, let y = 0.
0 = -x + 1
0 - 1 = -x
-1 = -x
1 = x

#3)
y = x + 1

m = +1, so the line rises from left to right.

y-intercept:
y = 0 + 1
y = 1

x-intercept:
0 = x + 1
0 - 1 = x
-1 = x

#2)
This is the equation of a parabola. A parabola can be written in the vertex form: y - k = a (x - h)², where (h, k) is the vertex of the parabola. In the given equation, y = x² - 4x, we can find out what h and k are by completing the trinomial square on the right side of the equation, and then expressing the entire equation in the vertex form given above. To complete the trinomial square, we first divide the coefficient of the x term by 2. Then we square that result and add it to the x² and linear x term. So h² = (-4/2)² = (-2)² = 4. Then, since we added 4 on the right side, either we must add 4 to the left side, or add -4 to the right side to keep the equation balanced. For your particular equation, for reasons which will become apparent shortly, I choose to do the latter:

y = x² - 4x + 4 - 4 ----> y = (x - 2)² - 4, which can be rewritten as y + 4 = (x - 2)². Now we have the equation in the vertex form for a parabola, y - k = a (x - h)², where a = 1, h = 2 and k = -4. Do you see why k = -4? Note that when x = h, (x - h) = (h - h) = 0 and (h - h)² = 0² = 0, and y = 0 + k = k. Hence, the assertion is made that (h, k) is the vertex of a parabola.

To tell the direction a parabola opens, look at the constant, a, before the expression (x - h)². If a > 0 the parabola opens up. If a < 0, the parabola opens down. Believe it or not, a parabola can also be written in the form x - k = a (y - h)², where y is the variable being squared. In that case, if a > 0, then the parabola opens to the right. If a < 0, then it opens to the left.

In your particular equation a = 1, and x is the variable being squared, so the parabola opens up. So, given everything we have found concerning your particular parabola, its vertex is located at (2, -4) and opens up.

Answer 2

In order to find the axis intercepts, you simply have to set each variable to equal 0.
For example, in number 1, you know that the y-intercept occurs when x is equal to zero. So, y = -x + 1 becomes y = -(0) + 1 and therefore y = 1 and the y-intercept is the point (0,1)
the x intercept occurs when y = 0, so y = -x + 1 becomes 0 = -x + 1 or x = 1. The x-intercept being the point (1,0)

To find the direction of a line, you should examine the equation in the point-slope formula of y = mx + b. In this formula, m stands for the slope, and b stands for the y-intercept.
A line with a positive slope will travel from the lower left to the upper right, like this: /
A line with a negative slope will travel from the upper left to the lower right like this: \

Examining number 1. in point slope formula looks like this:
y = (-1)x + 1 where m = -1 and b = 1. Therefore, we know the slope of the line is negative, and we can determine the direction of the graph.

Answer 3

isn't skool over?