A vertex and another point pin down a quadratic: the vertex fixes the shifts; the other point fixes the multiplier. Use vertex form, then a Desmos regression to find that multiplier. Expand if the choices use standard form. Watch the sign inside the square: puts the vertex at , not .
Hints
- Hint 1
A vertex is where a parabola turns. In , the squared term becomes zero when , so the vertex is . What numbers from the given vertex fill in and ?
- Hint 2
A point on means . The first coordinate is the input, not the output. What equation do you get when you put the other given point into vertex form?
- Hint 3
The only unknown left is the multiplier . A regression uses in place of to find it in Desmos. Then expand the square, keeping the inside sign straight, to match the choices.
Step-by-step
Build the quadratic from vertex form
Step 1Write a form with the given vertex
The vertex is where the parabola turns. In vertex form, , the square is zero at , so the vertex is . For , write: . The minus inside makes the square zero at . The multiplier still needs to be found.
- Step 2
Turn the other point into an equation
The graph passes through , so input must give output : . This point can determine . Using the vertex instead would make the squared term zero, so would disappear.
- Step 3
Find the multiplier in Desmos
Type . The asks Desmos to find the unknown multiplier that makes the point equation true. Under PARAMETERS, Desmos shows .
- Step 4
Put the multiplier into vertex form
Replace with : . The vertex stays at ; the other point determined which parabola with that vertex you need.
- Step 5
Expand to match the choices
To compare with equations in standard form, expand the square: . The middle term is , because multiplying by twice gives .
- Step 6
Combine the constants and choose
Combine the constants: . This equation has the given vertex and passes through the given point, so it defines . Choice D.