Write each equation in the form y = a(x-h)^2 + k.?
This process is known as 'completing the square', and it is a technique used to solve quadratic equations which do not factorise easily. It is used to derive the formula for solving quadratics. y = 2x^2 + 20x + 50 You want to get this into the form y = a(x-h)^2 + k. First factor out the coefficient of x^2: y = 2(x^2 + 10x + 25) Note that the expression in the brackets is (x+5)^2. You can therefore write y = 2(x + 5)^2. a = 2, and because the expression in the brackets is a perfect square, k = 0. If you want to show the zero value of k, write your answer as: y = 2(x + 5)^2 + 0.
How would you write an equation in y-k=a(x-h)^2 form with the x-intercepts -2 and 6, and the y-intercept -6?
The roots (X intercepts) being -2 and 6 mean that the equation must have the form y = S(x + 2)(x - 6) where S is some real scaling factor. Adding the constraint that the Y intercept is -6 gives: -6 = S(0 + 2)(0 - 6), hence S = 1/2 and our equation is y = 1/2 (x + 2)(x - 6) = 1/2 [ x^2 - 4x - 12 ] Now complete the square on the term in brackets: x^2 - 4x - 12 = x^2 - 4x + 4 - 4 - 12 = (x - 2)^2 - 16 Hence the equation is: y = 1/2 [ (x - 2)^2 - 16 ] (y + 8) = 1/2 (x - 2)^2 which was the final desired form. Note that this form of a quadratic equation is useful because it tells us directly that we've got a parabola opening upward (like a "U") whose vertex is at the point (2,-8).
F (x)= 3x^2+9x-11 Write equation in vertex form?
Hey there! Here's the answer. f(x)=3x^2+9x-11 --> y=3x^2+9x-11 --> y+11=3x^2+9x --> y+11=3(x^2+3x) --> y+11+3(3/2)^2=3(x^2+3x+(3/2)^2) --> y+11+27/4=3(x^2+3x+9/4) --> y+71/4=3(x+3/2)^2 --> y=3(x+3/2)^2-71/4. Completing the square for x^2+bx is x^2+bx+(b/2)^2. Now, we have the equation in vertex form. y=3(x+3/2)^2-71/4. Now, similarly, here's the general vertex form for the parabola. y=a(x-h)^2+k. Setting these equations together and solve, we get h=-3/2 and k=-11, making the vertex (-3/2,-71/4). We know that a=3. If a>0, the parabola is facing up, making the vertex a minimum point. Since 3>0, the vertex is the minimum point. In order to find the axis of symmetry, use the equation x=h. Since h=-3/2, the axis of symmetry is x=-3/2. Hope it helps!
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!!!
How do you write a quadratic function in vertex form?
You mean standard form?The standard form of a quadratic function is given by[math]y=a(x-h)^2+k[/math]Here, vertex[math]=(h,k)[/math]This is how to get it….Suppose the given quadratic function is [math]y=ax^2+bx+c[/math]Factor out [math]a[/math] from the first two terms. This gives [math]y=a\left(x^2-\dfrac{b}{a}x\right)+c[/math]Complete the square[math]y=a\left(x^2–2\cdot x\cdot \dfrac{b}{2a}+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}\right)+c[/math][math]\implies y=a\left(x-\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a}+c[/math][math]\implies y=a\left(x-\dfrac{b}{2a}\right)^2-\dfrac{b^2–4ac}{4a}[/math]Now, substitute [math]h=\dfrac{b}{2a}[/math] and [math]k=\dfrac{4ac-b^2}{4a}[/math]So, can also writevertex [math]= \left(\dfrac{b}{2a},-\dfrac{D}{4a}\right)[/math]where [math]D[/math] is the discriminant.
Please help me? Write this hyperbola equation in standard form?
9x^2-4y^2-36x+8y-4=0 We need to put it into the form: A(x-h)^2 - B(y-k)^2 = C or better yet the form (x-h)^2/a^2 + (y-k)^2/b^2 = 1 Geezah above made a small error in line six: 9x^2 - 36x -4y^2 +8y = 4 9( x^2 - 4x ) -4 (y^2 - 2y ) =4 9( x^2 - 4x +4 ) -4 (y^2 - 2y+1 ) =4+9*4-4*1 9( x-2)^2 -4 (y-1 )^2=36 divide both sides by 36 [(x-2)^2]/4 - [(y-1)^2]/9 = 1
Pre Calc: Write the equation in standard form y=x^2-10x+4?
Standard form of quadratic function is: y = a(x - h)^2 + k y = x^2 - 10x + 4 y = x^2 - 10x +(-10/2)^2 + 4 - (-10/2)^2 y = x^2 - 10x + 25 + 4 - 25 y = (x - 5)^2 - 21
Write the equation, in standard form, of the circle with radius 4 and center (-1, 0)?
The equation of a circle is as follows:[math](x-h)^2+(y-k)^2=r^2[/math]Here, [math](h,k)[/math] is the centre of the circle and [math]r[/math] is the radius.On comparing with [math](-1,0)[/math], we get,[math]h=-1[/math] and [math]k=0[/math] along with [math]r=4[/math].Substituting these values in the equation, we get,[math](x-(-1))^2+(y-0)^2=(4)^2[/math][math](x+1)^2+y^2=16[/math]This is the equation of the circle with centre [math](-1,0)[/math] and radius [math]4[/math] which can further be expressed as[math]x^2+2x+1+y^2=16[/math][math]x^2+y^2+2x-15=0[/math]Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
PreCalculus: Need help on writing the standard form of the equation of the parabola?
Hi, The formula for a parabola in vertex form is f(x) = a(x-h)² + k Since the vertex is (5,12), plug those values in for h and k, but not for "a". y = a(x-5)² +12 Now plug (7,15) into this for x and y, so you can solve for "a". 15 = a(7-5)² +12 15 = 4a + 12 3 = 4a a = 3/4 y = ¾(x-5)² +12 <==ANSWER I hope that helps!! :-)