Graphing a Basic Parabola

Graph $f(x) = x^2 - 4x + 3$ and identify all key features.

Identify the coefficients: $a = 1$, $b = -4$, $c = 3$ = Parabola opens upward (since $a > 0$)

Find the axis of symmetry: $x = -\frac{b}{2a} = -\frac{-4}{2(1)} = \frac{4}{2} = 2$ = Axis of symmetry: $x = 2$

Find the vertex: $f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$ = Vertex: $(2, -1)$

Find the y-intercept: $f(0) = (0)^2 - 4(0) + 3 = 3$ = Y-intercept: $(0, 3)$

Find the x-intercepts (roots): $x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3) = 0$ = X-intercepts: $(1, 0)$ and $(3, 0)$

Plot points and draw the parabola: Plot vertex $(2, -1)$, y-intercept $(0, 3)$, x-intercepts $(1, 0)$ and $(3, 0)$ = U-shaped curve opening upward

Answer: The parabola has vertex $(2, -1)$, axis of symmetry $x = 2$, opens upward, crosses the x-axis at $x = 1$ and $x = 3$, and crosses the y-axis at $y = 3$.

Parabola Opening Downward

Graph $f(x) = -x^2 + 2x + 8$ and find the maximum value.

Identify the coefficients: $a = -1$, $b = 2$, $c = 8$ = Parabola opens downward (since $a < 0$)

Find the axis of symmetry: $x = -\frac{b}{2a} = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1$ = Axis of symmetry: $x = 1$

Find the vertex (maximum point): $f(1) = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = 9$ = Vertex: $(1, 9)$ - this is the maximum

Find the y-intercept: $f(0) = -(0)^2 + 2(0) + 8 = 8$ = Y-intercept: $(0, 8)$

Find the x-intercepts: $-x^2 + 2x + 8 = 0 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0$ = X-intercepts: $(4, 0)$ and $(-2, 0)$

Answer: The parabola opens downward with vertex $(1, 9)$, which is the maximum value of the function. It crosses the x-axis at $x = -2$ and $x = 4$.

Using the Discriminant

Without graphing, determine how many x-intercepts $f(x) = x^2 - 6x + 9$ has.

Identify the coefficients: $a = 1$, $b = -6$, $c = 9$ = Ready to calculate discriminant

Calculate the discriminant: $\Delta = b^2 - 4ac = (-6)^2 - 4(1)(9) = 36 - 36 = 0$ = $\Delta = 0$

Interpret the discriminant: $\Delta = 0$ means exactly one x-intercept (the vertex touches the x-axis) = One x-intercept (repeated root)

Find the x-intercept: $x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$ = The parabola touches the x-axis at $(3, 0)$

Answer: The function has exactly one x-intercept at $x = 3$. The parabola touches the x-axis at its vertex - this is called a "double root."

Mistake: Forgetting the negative sign when calculating the axis of symmetry

Why: The formula $x = -\frac{b}{2a}$ has a negative sign that's easy to miss, especially when $b$ is already negative.

Correct: Always write the formula with the negative sign first: $x = -\frac{b}{2a}$. If $b = -4$ and $a = 1$, then $x = -\frac{-4}{2} = \frac{4}{2} = 2$.

Mistake: Confusing the vertex with the y-intercept

Why: The y-intercept $(0, c)$ is often easier to find, but it's not the vertex unless the axis of symmetry is $x = 0$.

Correct: The vertex is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$. Calculate the x-coordinate first, then substitute to find the y-coordinate.

Mistake: Thinking all parabolas open upward

Why: Many students forget that the sign of $a$ determines the direction.

Correct: Always check the sign of $a$ first: $a > 0$ opens up (minimum), $a < 0$ opens down (maximum).

Mistake: Assuming every parabola crosses the x-axis

Why: Not all quadratic functions have real roots. The discriminant determines this.

Correct: Check $\Delta = b^2 - 4ac$: if $\Delta > 0$, two x-intercepts; if $\Delta = 0$, one; if $\Delta < 0$, none.

Projectile Motion

When you throw a ball, its height over time follows a parabolic path due to gravity.

A ball thrown upward has height $h(t) = -5t^2 + 20t + 1.5$ meters after $t$ seconds. The vertex gives the maximum height, and the positive root gives when it lands.

Business Revenue

Companies use quadratic functions to model how price changes affect revenue.

If a company's revenue is $R(p) = -2p^2 + 100p$ where $p$ is the price in euros, the vertex shows the price that maximizes revenue.

Q: What's the difference between a quadratic function and a quadratic equation?

A: A quadratic function is $f(x) = ax^2 + bx + c$ and produces outputs for any input. A quadratic equation is $ax^2 + bx + c = 0$ and we solve it to find specific x-values (the roots).

Q: Why is it called a parabola?

A: The word comes from the Greek 'parabole' meaning 'comparison' or 'placing side by side.' It was named by the Greek mathematician Apollonius around 200 BCE when studying conic sections.

Q: Can a parabola have no x-intercepts?

A: Yes! When the discriminant $\Delta = b^2 - 4ac$ is negative, the parabola never crosses the x-axis. For example, $f(x) = x^2 + 1$ has its lowest point at $(0, 1)$, which is above the x-axis.