Parabolas: Additional Practice (Set 1)

Use the discriminant with $a = 1$, $b = 1$, $c = 1$.

$D = (1)^2 - 4(1)(1) = 1 - 4 = -3$

Since $D < 0$ there are no real solutions, so the parabola has no x-intercepts.

Half of $-2$ is $-1$, and $(-1)^2 = 1$, so add and subtract $1$.

$y = (x^2 - 2x + 1) - 1 - 8 = (x - 1)^2 - 9$

The square is $(x - 1)^2$ and the constants combine to $-1 - 8 = -9$.

The vertex lies on the axis of symmetry, halfway between the two x-intercepts. Average them.

$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$

So the vertex's x-coordinate is $5$.

The factored form shows the x-intercepts are $x = 2$ and $x = 6$, and the axis is halfway between them.

$x = \frac{2 + 6}{2} = 4$

So the axis of symmetry is $x = 4$.

Compute the discriminant with $a = 9$, $b = -12$, $c = 4$.

$D = (-12)^2 - 4(9)(4) = 144 - 144 = 0$

Since $D = 0$ there is exactly one x-intercept. Indeed $9x^2 - 12x + 4 = (3x - 2)^2$, touching the axis at $x = \tfrac{2}{3}$.

Vertex form $y = a(x - h)^2 + k$ needs $(h, k) = (-2, 5)$, so $x + 2$ inside and $+5$ outside, with $a < 0$ for downward.

$y = -(x + 2)^2 + 5$

The others fail: $(x + 2)^2 + 5$ opens upward; $-(x - 2)^2 + 5$ has vertex $(2, 5)$; $(x - 2)^2 - 5$ has vertex $(2, -5)$ and opens upward.

Compare with vertex form $y = a(x - h)^2 + k$: here $h = 1$ and $k = 12$.

$(h, k) = (1, 12)$

The leading coefficient $-3$ sets the direction and width but does not move the vertex.

The vertex x-coordinate is $-\tfrac{b}{2a} = -\tfrac{3}{2}$. Substitute it back for the height.

$y = \left(-\tfrac{3}{2}\right)^2 + 3\left(-\tfrac{3}{2}\right) + 2 = \tfrac{9}{4} - \tfrac{18}{4} + \tfrac{8}{4} = -\tfrac{1}{4}$

So the vertex is $\left(-\tfrac{3}{2}, -\tfrac{1}{4}\right)$.

Expand the square, distribute the $2$, and combine constants.

$y = 2(x^2 - 2x + 1) - 3 = 2x^2 - 4x + 2 - 3 = 2x^2 - 4x - 1$

So the standard form is $y = 2x^2 - 4x - 1$.

Set $y = 0$ and factor: two numbers that multiply to $5$ and add to $-6$ are $-1$ and $-5$.

$x^2 - 6x + 5 = (x - 1)(x - 5) = 0$

So $x = 1$ or $x = 5$.

The leading coefficient $a = -\tfrac{1}{2}$ is negative, so the parabola opens downward. With $b = 0$ the vertex sits on the y-axis, and setting $x = 0$ gives $y = 2$.

$a < 0 \;\Rightarrow\; \text{opens down}, \qquad (0, c) = (0, 2)$

So it opens down with vertex $(0, 2)$.

The parabola opens upward with vertex $(0, 4)$, already above the axis. The discriminant confirms it with $b = 0$, $c = 4$.

$D = 0^2 - 4(1)(4) = -16 < 0$

A negative discriminant means no real x-intercepts, so the graph meets the x-axis at zero points.