Sketch the graph of f(x) = x² + 3x − 4. Hence sketch y=f(x−2), y=f(x)+3, y=f(x+3), y=f(x), f(x-2)+3, y=(x+2)−3.

To sketch the graph of $f(x)$, we identify its key features:

• Shape: Since the coefficient of $x^2$ is positive ($a=1$), the parabola opens upwards.
• Vertex: The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. For $f(x) = x^2 + 3x - 4$, $a=1$ and $b=3$. $$x = -\frac{3}{2(1)} = -\frac{3}{2}$$ Substitute $x = -\frac{3}{2}$ into $f(x)$ to find the y-coordinate: $$f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) - 4 = \frac{9}{4} - \frac{9}{2} - 4 = \frac{9}{4} - \frac{18}{4} - \frac{16}{4} = \frac{9-18-16}{4} = -\frac{25}{4}$$ The vertex is at $\left(-\frac{3}{2}, -\frac{25}{4}\right)$.
• x-intercepts: Set $f(x) = 0$ and solve for $x$. $$x^2 + 3x - 4 = 0$$ Factor the quadratic equation: $$(x+4)(x-1) = 0$$ So, $x = -4$ or $x = 1$. The x-intercepts are $(-4, 0)$ and $(1, 0)$.
• y-intercept: Set $x = 0$. $$f(0) = (0)^2 + 3(0) - 4 = -4$$ The y-intercept is $(0, -4)$.

This transformation shifts the graph of $f(x)$ horizontally. Replacing $x$ with $(x-2)$ shifts the graph 2 units to the right.

This transformation shifts the graph of $f(x)$ vertically. Adding 3 to $f(x)$ shifts the graph 3 units upwards.

This transformation shifts the graph of $f(x)$ horizontally. Replacing $x$ with $(x+3)$ shifts the graph 3 units to the left.

This is the original graph of $f(x)$ as described in Step 1, with no transformation.

This transformation combines a horizontal and a vertical shift. Replacing $x$ with $(x-2)$ shifts the graph 2 units to the right, and adding 3 to the function shifts it 3 units upwards.

This transformation combines a horizontal and a vertical shift. Replacing $x$ with $(x+2)$ shifts the graph 2 units to the left, and subtracting 3 from the function shifts it 3 units downwards.

3 de faites, encore 2 aujourd'hui. Tu progresses.