How do you find the equation of a parabola in vertex & factored form?
Since you didn't include a link to a picture of the parabola, I can only give you a general answer. Remember the vertex form looks like this: y = a(x - h)² + k Look at the graph of the parabola. What are the coordinates of the vertex? Example: vertex is at (4, -7) Given: vertex at (4, -7) Means: h = 4, k = -7 Plug these values into the equation. y = a(x - 4)² + -7 y = a(x - 4)² - 7 Find another point on the parabola. I recommend using either an x-intercept or an y-intercept. Example: the parabola crosses the x axis at (5, 0) Given: point (5, 0) Means: x = 5, y = 0 Plug these values into the equation you have so far and solve for a. y = a(x - 4)² - 7 0 = a(5 - 4)² - 7 0 = a(1)² - 7 0 = a(1) - 7 0 = a - 7 a = 7 Update your equation. y = a(x - 4)² - 7 y = 7(x - 4)² - 7 <========= vertex form Expand the RHS. y = 7(x - 4)² - 7 y = 7[(x - 4)(x - 4)] - 7 y = 7[x(x) + x(-4) - 4(x) - 4(-4)] - 7 y = 7(x² - 4x - 4x + 16) - 7 y = 7(x² - 8x + 16) - 7 y = 7(x²) + 7(-8x) + 7(16) - 7 y = 7x² - 56x + 112 - 7 y = 7x² - 56x + 105 <===== expanded form Factor. y = (7x - 21)(x - 5) <==== factored form
Find the equation to the parabola while given the X and Y intercepts?
Greetings, y = a(x - 2)(x + 3), the general parabolic equation given the x-intercepts 5 = a(-2)(3), the y intercept a = -5/6 y = (-5/6)(x - 2)(x + 3) Regards
How do you find the equation of the parabola with intercepts: (-10.5,0); (0,3); (0,7)?
In the general case, you would write y = a x^2 + bx + c, (or rather, in this case x = a y^2 + b y + c) fill in the 3 points one by one and get a system of 3 equations in 3 unknowns. Since you are given intercepts, this problem is much easier than that. Consider y = 0, giving two intercepts. What must be factors of the polynomial in y? (There are two y intercepts so this is a parabola “on its side”.) Given those factors, the only thing left is the multiplying constant a, and you can fill in the other intercept to solve for a.
Equation of a parabola with x-intercepts of 6 and 18 and a y-intercept of 72?
In factored form: a(x-6)(x-18) = 0 y-intercept of 72, so a(0-6)(0-18) = 72, therefore a=2/3 y = (2/3)(x-6)(x-18) You could expand, then complete the square, but the x-coordinate of the vertex is in between the roots, i.e. x=12. When x=12, y=-24, so in vertex form, equation is: y = (2/3)(x-12)² - 24
How do I find the equation of a parabola given its zeros and a point?
To answer this question, I have to make certain assumptions. First, I am going to assume that you are talking about the graph of a quadratic function, and therefore a parabola that either opens up or down. Second, I am going to assume that you are given two real zeros, which represent the x-values of the points where the parabola crosses the x-axis (also called x-intercepts or roots). Third, I am going to assume that the point given is not the vertex of such parabola. With this information, now you can decide which of the following three forms of a quadratic equation you would use:1) y = ax^2 + bx + c2) y = a(x - h)^2 + k3) y = a(x - r₁)(x - r₂)Because you are given the two roots or zeros, the easiest equation to use is the 3rd one. You would want to use the 2nd one if you were given the vertex, and you would want to use the 1st one if you were given three random points. Let’s suppose that the roots or zeros given are x1 = -1, x2 = 4 , and the point given is (6, 7). So by substitution, you would first get the following:y = a(x - -1)(x - 4) or y = a(x + 1)(x - 4).Now substitute the x and y coordinates from the point given to solve for a:7 = a(6 + 1)(6 - 4) —-> 7 = a(7)(2) —-> 7 = 14a —-> thus, a = 0.5 or 1/2 .Consequently, your quadratic equation in root form will be…y = 0.5(x + 1)(x - 4) or y = 0.5*x^2 - 1.5x - 2 in expanded form.I hope this helped answer your question.
How do you find the x-intercept of a parabola?
Let's start from the beginning with a quadratic. Let y=f(x)=ax^2+bx+c, where a is not equal to zero. The y-intercept is the point where the parabola intersects the y-axis. So, the point is (0,f(0))=(0,c). If the parabola opens upward (a positive) and its vertex is above the x-axis or opens downward (a negative) and its vertex is below the x-axis, there aren't any (real) x-intercepts. Otherwise, if the quadratic has x-intercepts, solve ax^2+bx+c=0. Try factoring first and if that doesn't work easily, then try using the quadratic formula, x=(-b \pm \sqrt{b^2-4ac})/(2a). If a parabolar has x-intercepts it may have one or two so be sure to check your work carefully. Examples: y=x^2+1 has y-intecept (0,1) by no x-intercepts y=x^2-2x+1=(x-1)^2 has y-intercept (0,1) and x-intercept at (1,0) y=x^2-x-2=(x-2)(x+1) has y-intercept (0,-2) and x-intercepts (2,0) and (-1,0). A good idea: Given a quadratic either try to factor it so it looks like y=(a x+b)(c x+d) or try to write in the form y=a(x-b)^2+c. In the second case, (b,c) is the vertex and the sign of a tells you whether it goes up or down. The first case shows that the x-intercepts are x=-b/a and x=-d/c. Excellent question! :)
How do you write the equation of the parabola in vertex form?
The equation of parabola is given by :y^2 = 4 a x {If the axis of parabola is x-axis}…(i)and;x^2 = 4 a y {If the axis of parabola is y-axis}…(ii)in both the above equations {i.e, eq(i) and eq(i)}the vertex is Origin i.e, (0,0).if the vertex is (h,k) then the above written equations become:(y-k)^2 = 4 a (x-h) {If the axis of parabola is x-axis}…(iii)and(x-h)^2 = 4 a (y-k) {If the axis of parabola is y-axis}…(iv).So equation (iii) and equation (iv) are the required equations. Ans.I think it is a little bit difficult way . But if you try to realize that “what do the literals (alphabets) in this equation means and how the distance formula is used”, then it will be very easy for you to remember these equations.you can ask me if you don’t know about the literals of them.Thanks!!!
What do equations of parabolas that can't be factored tell you about their x-intercepts?
Notice that the quadratic and linear terms are identical in the three parabolas. If you plotted them all, they would be the same size and shape, but shifted vertically with respect to y = x² - 4x + 4. If you could only plot y = x² - 4x + 4, you could sketch the other two by transferring each point up t for y = x² - 4x + 6 and transferring each point down two for y = x² - 4x + 2. All three vertices lie on the same vertical line. If you know the vertex of y = x² - 4x +4, the other two vertices are located two above and two below.
A parabola has two x intercepts at (-2 , 0) and (3 , 0) and passes through the point (5 , 10). What is the equation of this parabola?
ax^2 + bx + c = 0———a(-2)^2 + b(-2) + c = 0 -> c = -4a + 2ba3^2 + b3 + c = 0 -> 9a + 3b -4a + 2b = 0 -> a = -ba5^2 + b5 + c = 10 -> 25a + 5b -4a + 2b = 10 -> -25b + 5b + 4b + 2b = 10 -> -14b = 10———b = -10/14 = -5/7a = 10/14 = 5/7c = -60/14 = -30/7———Answer: (5/7)x^2 - (5/7)x - 30/7
Parabola factored form help?
"Determine: • the equation (in factored form) of quadratic relation and • the direction of opening of the parabola." NOTE: There are other forms of the quadratic, like: y = a(x – p)² + q that show the vertex V(p, q) and also ... y = ax² + bx + c that show the y-intercept (0, c) (This is also called "Standard Form") y = a(x – r)(x – s) is just another form of the quadratic function that happens to display the x-intercepts. Given that the x-intercepts are –2, and 4 ∴r = 2, s = –4 y = a(x + 2)(x – 4), and now we just need to find "a" But we have one more piece of info: the y-intercept is 5 ∴(0, 5 ) is on the graph ∴ the point's co-ords satsify (ie makes true) the equation. So substitute the co-ords into the eq'n and solve for "a' 5 = a(0 +2)(0 – 4) = –8a a = –5/8 So y = –5/8(x + 2)(x – 4) is the required equation. and ∵ a < 0 (ie negative) the graph open downward.