Select the x-coordinate of the vertex of the parabola defined in the function f(x)= -5x^2+3x+5?
Re-write f(x) into vertex form which is y = a(x - h)^2 + k where (h, k) is vertex: f(x) = -5x^2 + 3x + 5 = -5(x^2 - 3x/5 - 1) = -5((x - 3/10)^2 - 109/100) = -5(x - 3/10)^2 + 109/20 Therefore x coordinate of vertex is x = 3/10 Option 3
Select the x-coordinate of the vertex of the parabola defined by the function f(x) = -7x2 + 3x + 2.?
If easy, the equation of a parabola should invariably be converted into vertex form. If the equation is in vertex form, you can easily find the coordinates of the vertex. Equation in Vertex Form is y = (x-h)²+c and Standard Form is y = ax²+bx+c If you require only the x-coordinate of the Vertex when the equation is in the standard form, it is too easy. x-coordinate fo the vertes is -b/2a Here in your equation a = -7, b = 3 and c = 2 x-coordinate of the vertex is = -b/2a = -3/2×-7 = -3/-14 or 3/14 y-coordinate of the vertex is given by y = c-(b²/4a²) And the axis of the parabola which is also the line of symmetry is given by x = -b/2a If the equation of the parabola is in Vertex Form i.e. y = (x-h)²+k Then the coordinates of the Vertex are (h,k) and equation of the line of symmetry is x=h
Select the x-coordinate of the vertex of the parabola defined by the function f(x)-7+9x+6 a -7/9 b06 c9/14 d0-?
Needs some clarification. Edit: if you meant: f(x) = -7x^2 + 9x + 6, then: x = -9/(-2 * 7) = 9/14 c) => answer
Select the x-coordinate of the vertex of the parabola defined by the function f(x) = -3x^2 +6x+5?
f(x) = -3(x^2 - 2x) + 5 f(x) = -3(x^2 - 2x + 1 - 1) + 5 f(x) = -3(x^2 - 2x + 1) + 3 + 5 f(x) = -3(x - 1)^2 + 8 Answer: 1
Select the x-coordinate of the vertex of the parabola defined by the function?
y = ax^2 + bx + c is the parabola y = 9x^2+5x+2 x-coordinate of vertex is -b/2a = -(5) / (2)(9) = -5 / (2)(9) = -5/18 Method 2: y = 9x^2+5x+2 y-2 = 9x^2+5x y-2 = 9(x^2+ (5/9) x) y-2 = 9( x^2 + 2(5/18) x + (5/18)^2 - (5/18)^2 ) y-2 = 9( (x+5/18)^2 - 25/324) y- 2 = 9 (x+5/18)^2 - 225/324 y = 9(x+5/18)^2 + (2-225/324) y = 9(x+51/8)^2 + 47/36 x-coordiante of vertex is -5/18
Select the x-coordinate of the vertex of the parabola defined by the function f(x)=-3x^2+5x+9?
Let's complete the square: f(x) = -3x² + 5x + 9 f(x) = -3(x - 5x/3) + 9 f(x) = -3(x - 5x/3 + 25/36) + 9 - 3(-25/36) f(x) = -3(x - 5/6)² + 133/12 Here, the x-coordinate is the the number inside the bracket, but we give it the opposite sign. That means our x-coordinate is + 5/6
Select the x-coordinate of the vertex of the parabola defined by the function f(x) = -9x^2 + 7x + 9.?
y = -9x² + 7x + 4 Put the equation into vertex form y = a(x - h)² + k factor out leading coefficient y = -9(x² - (7/9)x) + 4 complete the square y = -9(x² - (7/9)x + (7/18)²) + 9(7/18)² + 4 y = -9(x - 7/18)² + 193/36 vertex (7/18, 193/36)
Find the coordinates of the vertex for the parabola defined by the given quadratic function f(x)=3x²+12x+4?
You can use the vertex formula (-b/2a,f(-b/2a)) The x-coordinate of the vertex will be -b/2a The y-coordinate of the vertex will be f(-b/2a).....meaning that once you find the x-coordinate, you substitute that number for all the x's in 3x²+12x+4 If that doesn't make sense, remember that f(x) = ax²+ bx + c is the definition of a quadratic formula. x-coordinate of the vertex for 3x²+12x+4 -b/2a -12/2(3) =-2 y-coordinate of the vertex for 3x²+12x+4 SUBSTITUTE X-COORDINATE INTO FUNCTION f(-b/2a) 3(-2)²+12(-2) + 4 = -8 Vertex is (-2,-8)
What is the x-coordinate of the vertex of the parabola defined by the function f(x) = -7x^2 + 5x + 9?
The vertex formula is given by: ((-b/2a), (4ac - b^2)/4a) So the x-coordinate of the vertex is found by (-b/2a). Quadratic equation is in the form: f (x) = Ax^2 + Bx + C In this case, A = -7, B = 5 and C = 9, plug these numbers into the vertex formula... (-b/2a) = (-5/2(-7)) = -5/-14 = 5/14 So the x-coordinate of the vertex is 5/14.
How do I write in vertex and standard form a parabola with a vertex (-2,-5) that passes through (6, 635)?
Well, I don’t know how qualified I actually am, as I’m a Freshman taking Junior level math, but I’ll take a shot at it. I do appreciate pointing out any ateas where I messed up.Sooooo, let’s start out with basic vertex form equation (mainly because I actually understand vertex form).The form is y=a(x-h)^2+kAlso, sorry for the bad exponents, I am using a phone in my defenseFirst, you have to find h and k. The one inside the parenthesis (h) is equally to the x coordinate of the vertex, so we get -2=h.Then there is k, which is equal to the y of the vertex, so we get -5=k.So, we end up with this, y=a(x-(-2))^2+(-5)Doing simple algebra, this simplifies to y=a(x+2)^2–5.Now look at the remaining variables, a, x, and y. Well, we can put in numbers for our x and y by choosing a point on the line. In your case, (6,635). So let’s do it635=a(6+2)^2–5635=a(8)^2–5635=64a-5640=64a10=aThere you have it, your vertex form solved, y=10(x+2)^2–5.If you are confused wuth how that algebra plays out, feel free to ask me to elaborate.Now this is the part I am unsure aboutFrom my uneducated point of view of standard form’s relation to vertex form is very similar to polynomials.Ex: (2x+4)^2=4x^2+16x+16See where I am coming from? Well, here is my best educated guess. First, turn our equation into a poynomial.y=10((x+2)×(x+2))-5Now, it’s very hard to understand the steps if you have no clue what I am doing, because it works a lot better with paper! I’ll finish it and can elaborate better later.(x+2)(x+2)x^2+4x+4Now put this back into the equation and simplify.y=10(x^2+4x+4)-5y=10x^2+40x+35I believe that is it in standard form, but I have no clue as I have 0 knowledge of standard form for parabolas. Feel free to correct me, I am open to critizism. I hope I helped!