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Suppose the equation of the parabola is y = ax2 + bx + c
Now, where the parabola crosses the x-axis (the x intercepts), the value of y must be zero (that is what crossing the x-axis means). If the discriminant, b2 - 4ac is less than zero, y has no real roots. This means that there is no real value of x for which y equals zero and so the parabola has no x intercepts. If the discriminant is zero then the parabola only touches the x-axis - at (-b/2a,0). If the discriminant is greater than zero, there are two distinct intercepts.
If a>0 then the parabola is shaped like a U and is wholly above the x-axis. If a<0 then the parabola is an upturned U, wholly below the x axis. If a = 0 the quadratic term disappears and the function is a straight line, not a parabola.
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How do you find the equation of a parabola in standard form with x intercepts at 0 and 0 passing through 7 and 8?
y = 8/49*x2
How do you find the x-intercepts of this function parabola?
Solve the quadratic equation.If y = ax^2+bx+c then the intercepts arex = [-b +/- sqrt(b^2 - 4ac)]/(2a).The solutions are real if and only if b^2 >= 4ac
How do you graph quadratic functions in vertex form?
The standard form of quadratic function is: f(x) = a(x - h)^2 + k, a is different than 0 The graph of f is a parabola whose vertex it is the point (h, k). If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Furthermore, if |a| is small, the parabola opens more flatly than if |a| is large. It is a general procedure for graphing parabolas whose equations are in standard form: Example 1: Graph the the quadratic function f(x) = -2(x - 3)^2 + 8 Solution: Standard form: f(x) = a(x - h)^2 + k Given function: f(x) = -2(x - 3) + 8 From the give function we have: a= -2; h= 3; k = 8 Step 1. Determine how the parabola opens. Note that a = -2. Since a < 0, the parabola is open downward. Step 2. Find the vertex. The vertex of parabola is at (h, k). because h = 3 and k = 8, the parabola has its vertex at (3, 8). Step 3. Find the x-intercepts by solving f(x) = 0. Replace f(x) with 0 at f(x) = -2(x - 3)^2 + 8 and solve for x 0 = -2(x - 3)^2 + 8 2(x - 3)^2 = 8 (x- 3)^2 = 4 x - 3 = square radical 4 x - 3 = 2 or x -3 = -2 x = 5 or x = 1 The x- intercepts are 1 and 5. Thus the parabola passes through the points (1, 0) and (5, 0), this means that parabola intercepts the x-axis at 1 and 5. Step 4. Find the y-intercept by computing f(0). Replace x with 0 in f(x) = _2(x - 3)^2 + 8 f(0) = -2(0 - 3)^2 + 8 f(0) = -2(9) + 8 f(0) = -10 The y-intercept is -10. Thus the parabola passes through the point (0, -10), this means that parabola intercepts the y-axis at -10. Step 5. Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at -10. The axis of symmetry is the vertical line whose equation is x = 3. Example 2: Graphing a quadratic function in the form f(x) = ax^2 + bx + c Graph the quadratic function f(x) = -x^2 - 2x + 1 Solution: Here a = -1, b = -2, and c = 1 Step 1. Determine how the parabola opens. Since a = 1, a < 0, the parabola opens downward. Step 2. Find the vertex. We know that x-coordinate of the vertex is x = -b/2a. Substitute a with -1 and b with -2 into the equation for the x-coordinate: x = - b/2a x= -(-2)/(2)(-1) x = -1, so the x-coordinate of the vertex is -1, and the y-coordinate of the vertex will be f(-1). thus the vertex is at ( -1, f(-1) ) f(x) = -x^2 - 2x +1 f(-1) = -(-1)^2 - 2(-1) + 1 f(-1) = -1 + 2 + 1 f(-1) = 2 So the vertex of the parabola is (-1, 2) Step 3. Find the x-intercepts by solving f(x) = o f(x) = -x^2 -2x + 1 0 = -x^2- 2x + 1 We can't solve this equation by factoring, so we use the quadratic formula to solve it. we get to solution: One solution is x = -2.4 and the other solution is 0.4 (approximately). Thus the x-intercepts are approximately -2.4 and 0.4. The parabola passes through ( -2.4, 0) and (0.4, 0) Step 4. Find the y-intercept by computing f(0). f(x) = -x^2 - 2x + 1 f(0) = -(0)^2 - 2(0) + 1 f(0) = 1 The y-intercept is 1. The parabola passes through (0, 1). Step 5. graph the parabola with vertex at (-1, 2), x-intercepts approximately at -2.4 and 0.4, and y -intercept at 1. The line of symmetry is the vertical line with equation x= -1.
What is an example of an x -intercept?
An x-intercept is the point where a function intersects the x-axis on a Cartesian coordinate plane. For example, if the graph of a parabola is plotted and the graph intersects the x-axis on the coordinate plane, the point(s) where the graph intersects the x-axis are the x-intercepts for that function.
What is the graph of a quadratic formula?
In general, quadratic equations have graphs that are parabolas. The quadratic formula tells us how to find the roots of a quadratic equations. If those roots are real, they are the x intercepts of the parabola.
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 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)
What are the zeros of functions and what do they represent?
The zeros of functions are the solutions of the functions when finding where a parabola intercepts the x-axis, hence the other names: roots and x-intercepts.
How do you get the equation of of a parabola from its x-intercepts?
Let us say you X intercepts are -2 and 3 set up (X + 2)(X - 3) FOIL X^2 - X - 6 = 0 ----------------------- your parabolic equation
How do you find how many x intercepts the parabola has using the discriminant?
If the discriminant is negative, there are 0 interceptsIf the discriminant is zero, there is 1 interceptIf the discriminant is positive, there are 2 intercepts
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.
What are the roots of a parabola?
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
Why is factoring a useful tool when graphing a parabola?
Factoring will show you where the parabola intercepts the axis.
Where does the parabola x2 plus 8x equals 20 have roots?
x^2+8x=20 x^2+8x-20=0 (x+10)(x-2)=0 x=-10 and x=2 are the roots (intercepts) (-10,0) and (2,0) are the x-intercepts.
What are the characteristics of parabolas?
there are three main characteristics of a parabola. these are: 1. vertex: the point at the apex of a parabola 2. x- intercepts: the points at which the parabola intersects or touches the x axis. 3. face: if the parabola is in the form of the letter "u" then it's face is upwards. if the parabola is the in form of the inverted letter "u" then it face downwards :D
How do you find the x-intercept of a parabola?
Factorise equation, and look at what x values are needed for the equation to equal zero. Eg. x^2+5x+6 (x+3)(x+2)=0 So parabola intercepts x axis at -3 and -2.
What is the maximum number and the minimum number of intercepts for a parabola?
there can be 0, 1, 2, or infinite intercepts for a parabola 0 intercepts occurs when the parabola does not meet the 2nd line 1 occurs when the parabola intersects a line at the vertex 2 occurs when the line does not intersect at the vertex, but still intersects the parabola infinite occurs when there are 2 parabolas, that although they may be written differently, are the same on a graph.
