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When the graph of a quadratic function crosses the x-axis twice the x-coordinate of the vertex lies between the two x-intercepts?
The x co-ordinate of a quadratic lies exactly halfway between the two x-intercepts, assuming they exist. Alternatively, the x co-ordinate can be found using the formula -B/(2A), when the function is in the form, y = Axx + Bx + C.
What represents the vertex and two x-intercepts of y equals x2-2x-35?
(1,-36); -5 and 7y = x2-2x-35; f(x) = ax2+bx+cvertex, method 1:There is a formula for the vertex, if you can remember it. (h,k) = [-b/2a, f(-b/2a)].-b/2a = 2/2x1 = 2/2 = 1. f(1) = 12-2x1-35 =1-2-35 = -36. The vertex is (1,-36).vertex, method 2:Otherwise you complete the square to convert to the vertex form, y = a(x-h)2 + k. Complete the square for x2-2x: the constant term will be (-2/2)2 = (-1)2 = 1.y = (x2-2x+1)-1-35 = (x-1)2 -36 ==> The vertex is (h,k) = (1,-36).x-intercepts, method 1:To find the x-intercepts, let y=0: 0 = x2-2x-35 (x-7)(x+5) = 0 x-7=0 or x+5=0 x=7 or x=-5. The intercepts are -5 and 7.x-intercepts, method 2:There is also a nice trick to find the intercepts: 36 = 62. the intercepts are 1+/- 6 = 7 and -5.
What is a term used to describe the relationship between two variables who's graph is a straight line passing through the point 00?
It is a straight line equation with no x or y intercepts on the Cartesian plane
What is the parabola?
A parabola is a type of graph that is not linear, and mostly curved. A parabola has the "x squared" sign in it's equation. A parabola is not only curved, but all the symmetrical. The symmetrical point, the middle of the parabola is called the vertex. You can graph this graph with the vertex, x-intercepts and a y-intercept. A parabola that has a positive x squared would be a smile parabola, and the one with the negative x squared would be a frown parabola. Also, there are the parabolas that are not up or down, but sideways Those parabolas have x=y squared, instead of y = x squared.
How are solution and x-intercept and zero of a function and and root related?
solutions are the well solution to the problem. X-intercepts are wherever a graph cross the x axis, which are hte solutions when you have to find out what x is, zeros are pretty much the same thing although i think that include y-intercepts as well..... not sure. and roots are the same thing as x-intercepts. so they are all more or less the same thing
How would you use intercepts to find the vertex in a quadratic equation with two x intercepts?
The vertex must be half way between the two x intercepts
How do you find the x intercepts using the vertex and y intercept?
It depends on the vertex of what!
What is the vertex of a parabola?
the vertex of a parabola is the 2 x-intercepts times-ed and then divided by two (if there is only 1 x-intercept then that is the vertex)
When the graph of a quadratic function crosses the x axis twice the x coordinate of the vertex lies between the two x intercepts?
that's true
When the graph of a quadriatic function crosses the x axis twice the x cordinate of the vertex lies where between the two intercepts?
Precisely midway. That is to say, at their mean (average).
When the graph of a quadratic function crosses the x-axis twice the x-coordinate of the vertex lies between the two x-intercepts?
The x co-ordinate of a quadratic lies exactly halfway between the two x-intercepts, assuming they exist. Alternatively, the x co-ordinate can be found using the formula -B/(2A), when the function is in the form, y = Axx + Bx + C.
What represents the vertex and two x-intercepts of y equals x2-2x-35?
(1,-36); -5 and 7y = x2-2x-35; f(x) = ax2+bx+cvertex, method 1:There is a formula for the vertex, if you can remember it. (h,k) = [-b/2a, f(-b/2a)].-b/2a = 2/2x1 = 2/2 = 1. f(1) = 12-2x1-35 =1-2-35 = -36. The vertex is (1,-36).vertex, method 2:Otherwise you complete the square to convert to the vertex form, y = a(x-h)2 + k. Complete the square for x2-2x: the constant term will be (-2/2)2 = (-1)2 = 1.y = (x2-2x+1)-1-35 = (x-1)2 -36 ==> The vertex is (h,k) = (1,-36).x-intercepts, method 1:To find the x-intercepts, let y=0: 0 = x2-2x-35 (x-7)(x+5) = 0 x-7=0 or x+5=0 x=7 or x=-5. The intercepts are -5 and 7.x-intercepts, method 2:There is also a nice trick to find the intercepts: 36 = 62. the intercepts are 1+/- 6 = 7 and -5.
Does a parabola always have 2 x intercepts?
No, if the vertex of the parabola is (0, 0) it will only have one x intercept. The parabola might have zero x intercepts as well. For example: Y= x^2 + 1 would never touch the x line.
What is X squared plus six x plus five?
It is the parabolic function defined by the product (x+1)(x+5), with x intercepts at -1 and -5, y-intercept at (0,5) and a vertex at (-3, -4).
How do you find the axis of symmetry and vertex of y equals x squared plus 6x plus 10?
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
How do you find the equation of a parabola with x intercepts -1 and 3 and y intercept 8?
Steps1) Find the co-ordinates of the vertex, by adding the x-intercepts and dividing by 2.Xv= (-1+3) / 2This gives you the x-coordinate of the vertex, 1. The vertex of the parabola is at (1, 8)2) Sub into the vertex form equation. (y= a(x-h)2+k h = x coord of vertex, k = y coordy=a(x-1)2+8note: when subbing in x value, always put the reverse sign into the equation, if positive, sub it in as negative, if negative, sub in as positiveNow you can't solve for a without knowing what point is passes through on the y-axis. after you know that you just solve sub x and y in, solve for a, then put a into the equationex: y= 2(x-1)2+8
What is the relationship between the deggre number of zeros or x intercepts on a graph?
I am not really sure what you are asking for, but any intercept on the x-axis has a y value of 0, so for any particular x value N, the intercept is at (N, 0).
