Sketch the parabola described by the equation 𝑦 = 2𝑥^2 − 20𝑥 + 42.

Answer 1

To sketch the parabola described by the equation 𝑦 = 2𝑥^2 − 20𝑥 + 42, we can first find the vertex. The x-coordinate of the vertex is given by 𝑥 = −𝑏/2𝑎, where 𝑏 is the coefficient of 𝑥 and 𝑎 is the coefficient of 𝑥^2. In this case, 𝑏 = −20 and 𝑎 = 2, so 𝑥 = −(−20)/(2×2) = 5.

Substituting 𝑥 = 5 into the equation, we find the y-coordinate of the vertex: 𝑦 = 2×5^2 − 20×5 + 42 = 2.

Therefore, the vertex of the parabola is (5, 2).

Next, we can find the y-intercept by setting 𝑥 = 0: 𝑦 = 2×0^2 − 20×0 + 42 = 42. The y-intercept is (0, 42).

To find the x-intercepts, we can solve the equation 𝑦 = 2𝑥^2 − 20𝑥 + 42 for 𝑥 when 𝑦 = 0:

0 = 2𝑥^2 − 20𝑥 + 42

Dividing both sides by 2:

0 = 𝑥^2 − 10𝑥 + 21

Factoring:

0 = (𝑥 − 3)(𝑥 − 7)

Therefore, the x-intercepts are (3, 0) and (7, 0).

We can now sketch the parabola using these points:

- The vertex is at (5, 2).
- The y-intercept is at (0, 42).
- The x-intercepts are at (3, 0) and (7, 0).

The parabola opens upward because the coefficient of 𝑥^2 is positive. The graph should look something like this: