Assume the expression is:
y = x² - 10x + 30
Complete the squares to get:
y = x² - 10x + 25 + 30 - 25
= (x - 5)² + 5
So the expression is in vertex form y = (x - h)² + k
The question does not contain an equation: only an expression. An expression cannot have a vertex form.
The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.
please help
To determine the equation of a parabola with a vertex at the point (5, -3), we can use the vertex form of a parabola's equation: (y = a(x - h)^2 + k), where (h, k) is the vertex. Substituting in the vertex coordinates, we have (y = a(x - 5)^2 - 3). The value of "a" will determine the direction and width of the parabola, but any equation in this form with varying "a" values could represent the parabola.
Vertex form is denoted by: y=a(x-h)2+k Where (h,k) is the vertex. So, we have: y=a(x-2)2+3 (This super\subscript thing is annoying). Plug in the values for x and y for the point in the equation and you have your answer.
The vertex form for a quadratic equation is y=a(x-h)^2+k.
2
The question does not contain an equation: only an expression. An expression cannot have a vertex form.
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.
The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.
-2
please help
To find the vertex of a quadratic equation in standard form, (y = ax^2 + bx + c), you can use the vertex formula. The x-coordinate of the vertex is given by (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))).
y=2(x-3)+1
y=x2-10x+30=(x-5)2-25+30=(x-5)2+5
look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)