How do you convert vertex form to standard form?

Square the bracket, multiply by a, then add k. For (x + 2)² + 5 that is x² + 4x + 4 + 5 = x² + 4x + 9. In general a(x - h)² + k = ax² - 2ahx + ah² + k, so b = -2ah and c = ah² + k, and the page shows both lines with your values substituted.

How do I find the zeros or the factored form from vertex form?

Press Solve this one and the solver below gives them: 2(x - 3)² - 8 has the roots 5 and 1 and the factored form 2(x - 5)(x - 1). By hand, set the vertex form to zero: 2(x - 3)² = 8, so (x - 3)² = 4 and x = 3 ± 2. When a and k have the same sign, (x - h)² would have to be negative, so there are no real zeros and the solver gives the complex pair instead.

Where is the y-intercept?

At (0, c), and c = ah² + k is the constant the expansion works out. For 2(x - 3)² - 8 it is 2(3)² - 8 = 10, so the curve crosses the y axis at (0, 10). The solver's working names it alongside the x-intercepts.

What if h or k is a fraction or a decimal?

Fractions like 1/2, decimals and negative values are all read, and b and c are worked out exactly as fractions, so 3(x - 1/3)² gives 3x² - 2x + 1/3. When b or c is a fraction, Solve this one multiplies every term by the common denominator before sending it, here 9x² - 6x + 1, which has the same roots in whole numbers, so the solver gives the double root exactly as 1/3.

Vertex Form to Standard Form Calculator

Type a, h and k from y = a(x - h)² + k and the standard form appears at once, with b = -2ah and c = ah² + k worked out in your numbers. This page opens on 2(x - 3)² - 8, which multiplies out to 2x² - 12x + 10, and the solver below opens on that same equation. Solve this one sends a, b and c to the quadratic solver below, which in Solve for x gives the roots, the factored form, the intercepts and a drawing of the parabola, so a vertex form reaches factored or intercept form in two steps.

Start from vertex form

Type a, h and k from y = a(x - h)² + k to multiply it out into standard form y = ax² + bx + c. Solve this one sends the result to the solver for its roots, factors and graph.

a
h
k

y = 2x² - 12x + 10

Start from y = 2(x - 3)² - 8, so a = 2, h = 3 and k = -8.
Square the bracket: (x - h)² = x² - 2hx + h², so a(x - h)² + k = ax² - 2ahx + ah² + k.
b = -2ah = -2(2)(3) = -12
c = ah² + k = 2(3)² - 8 = 10
Standard form: y = 2x² - 12x + 10

Rootsx = 5, 1a=2 b=-12 c=10 discriminant 64

A parabola opening upward, vertex at (3, -8). It crosses the x axis at 5 and 1.

a
b
c
Give me

x₁: 5
x₂: 1
Discriminant: 64
Form: Real, distinct

The roots of ax² + bx + c = 0, the discriminant that decides how many there are, and each root put back into the equation.

The working

x = (-b ± √(b² - 4ac)) / (2a), with a = 2, b = -12, c = 10
D = b² - 4ac = (-12)² - 4(2)(10) = 64
√D = 8, so x = 5, 1
Both roots come from q = -(b + sign(b)√D)/2, read as q/a and c/q. That is the stable form: written as (-b ± √D)/(2a), the smaller root loses digits whenever b² is far larger than 4ac.
h = -b/(2a) = 12/4 = 3
k = c - b²/(4a) = 10 - 144/8 = -8
Vertex form: 2(x - 3)² - 8, vertex (3, -8), axis of symmetry x = 3
Factored form: 2x² - 12x + 10 = 2(x - 5)(x - 1)
Intercepts: the curve crosses the x axis at (5, 0) and (1, 0), and the y axis at (0, 10).
Exact: a, b and c are whole numbers, so the discriminant is exact and the roots are given as fractions or surds beside their decimals.

Each root, put back into the equation

A root is only a root if it makes the left-hand side zero. The residual is the left-hand side worked out at the root this page computed, so one marked as rounding is the last-digit error of double arithmetic and not a second answer.

x₁ = 5: 2(5)² - 12(5) + 10 = 0
x₂ = 1: 2(1)² - 12(1) + 10 = 0

Three equations worth trying

One of each kind the discriminant allows. Loading one keeps the mode you are in.

a=1 b=1 c=-6D = 25
x = 2, -3
a=1 b=2 c=1D = 0
x = -1 (double)
a=1 b=0 c=1D = -4
x = ±i

The discriminant decides how many roots there are. b² - 4ac above zero gives two real roots, exactly zero gives one repeated root, and below zero gives a conjugate pair with an imaginary part. That is why it sits in the readout next to the answer rather than buried in the working: it is the number that says which of the three answers you are looking at. With a at zero there is no x² term at all, so the equation is linear and is solved as one.

Common questions

How do you convert vertex form to standard form?: Square the bracket, multiply by a, then add k. For (x + 2)² + 5 that is x² + 4x + 4 + 5 = x² + 4x + 9. In general a(x - h)² + k = ax² - 2ahx + ah² + k, so b = -2ah and c = ah² + k, and the page shows both lines with your values substituted.
How do I find the zeros or the factored form from vertex form?: Press Solve this one and the solver below gives them: 2(x - 3)² - 8 has the roots 5 and 1 and the factored form 2(x - 5)(x - 1). By hand, set the vertex form to zero: 2(x - 3)² = 8, so (x - 3)² = 4 and x = 3 ± 2. When a and k have the same sign, (x - h)² would have to be negative, so there are no real zeros and the solver gives the complex pair instead.
Where is the y-intercept?: At (0, c), and c = ah² + k is the constant the expansion works out. For 2(x - 3)² - 8 it is 2(3)² - 8 = 10, so the curve crosses the y axis at (0, 10). The solver's working names it alongside the x-intercepts.
What if h or k is a fraction or a decimal?: Fractions like 1/2, decimals and negative values are all read, and b and c are worked out exactly as fractions, so 3(x - 1/3)² gives 3x² - 2x + 1/3. When b or c is a fraction, Solve this one multiplies every term by the common denominator before sending it, here 9x² - 6x + 1, which has the same roots in whole numbers, so the solver gives the double root exactly as 1/3.

Every answer shows its discriminant, and each root is substituted back so you can see the residual. Whole-number coefficients up to 10 million give exact results: rational roots as fractions, irrational and complex roots in surd form, each with a decimal beside it; other coefficients are solved in double precision, where the last digits can round. With a at zero this is not a quadratic: it is solved as a linear equation, and vertex form is declined.