Question

which graph shows $y = x^2 + 3x - 4$? (with three graph options, each with coordinate axes and a parabola)

Explanation:

Step1: Find y-intercept

Set \( x = 0 \) in \( y = x^2 + 3x - 4 \). Then \( y = 0 + 0 - 4 = -4 \). So the graph should cross the y-axis at \( (0, -4) \).

Step2: Find x-intercepts

Set \( y = 0 \), solve \( x^2 + 3x - 4 = 0 \). Factor: \( (x + 4)(x - 1) = 0 \). So \( x = -4 \) or \( x = 1 \). The graph should cross the x-axis at \( (-4, 0) \) and \( (1, 0) \).

Step3: Analyze the vertex (optional but helpful)

The x-coordinate of the vertex of \( y = ax^2 + bx + c \) is \( -\frac{b}{2a} \). Here, \( a = 1 \), \( b = 3 \), so \( x = -\frac{3}{2(1)} = -1.5 \). Plugging back, \( y = (-1.5)^2 + 3(-1.5) - 4 = 2.25 - 4.5 - 4 = -6.25 \). So vertex at \( (-1.5, -6.25) \).
Now check the graphs:

• First graph: y-intercept? Doesn't seem to be -4. Eliminate.
• Second graph: y-intercept at -4, x-intercepts at -4 and 1 (matches our findings), vertex around -1.5, -6.25. This fits.
• Third graph: y-intercept positive? Doesn't match y-intercept -4. Eliminate.

Answer:

The Middle Graph