Question

Determine the equation of the axis of symmetry.

Answer 1

To simplify keyboarding, I will let f(x) ≡ y.
Q1. y = -2(x-1)^2 +3,
so: y = -2x² + 4x + 1, upon expansion & tidying up.

The curve is a parabola that points 'upwards' - because the coefficient of x² is < 0.
The axis of symmetry - which will be parallel to the y-axis - will be perpendicular to the tangent at the point where y is a maximum. To find that maximum, we must find dy/dx and equate it to zero.

dy/dx = -4x + 4
For dy/dx = 0, we have 4x = 4. Thus x = + 1
► Hence, the parabola is symmetrical about the line x = + 1

When x = 1, we have:
y = -2(1)² + 4(1) + 1
So:
y = -2 + 4 + 1 = + 3
► Hence, the vertex is at (1,3)

The curve intercepts the x-axis when y = 0.
That is when:
-2x² + 4x + 1 = 0
Which can be re-written as:
2x² - 4x - 1 = 0.

Since this quadratic equation doesn't have any rational factors, we must resort to the general solution for a quadratic equation (which I assume you are familiar with).
It provides the following two values:
x = 1 + √6 / 2; x = 1 - √6 / 2.
► These are the values of where the curve intercepts the x-axis.

Since we have:
y = -2x² + 4x + 1,
when the curve intercepts the y-axis, x = 0.
► Hence, y = + 1: the intercept on the y-axis.

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► Your second Q. is solved in exactly the same way, but note that since the coefficient of x² is > 0, it will be a 'downwards' pointing parabola