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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.
What else can I help you with?
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.
How many x- intercepts can a parabola have?
A parabola can have zero, one, or two x-intercepts. If the parabola opens upwards and the vertex is above the x-axis, it will have no x-intercepts. If the vertex touches the x-axis, there is one x-intercept (a double root), and if it opens upwards or downwards and intersects the x-axis at two points, there are two x-intercepts. The number of x-intercepts is determined by the discriminant of the quadratic equation representing 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
What happens if a parabola has no x intercepts?
If a parabola has no x-intercepts, it means that its graph does not intersect the x-axis. This occurs when the value of the quadratic's discriminant (b² - 4ac) is less than zero, indicating that the quadratic equation has no real solutions. Consequently, the parabola opens either entirely above or entirely below the x-axis, depending on the sign of the leading coefficient. If the leading coefficient is positive, the parabola opens upwards; if negative, it opens downwards.
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.
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.
