Question

Find the equation of the graph shown below.

• A. \(y = 3x - 4\)
• B. \(y = 2x^2 - 40\)
• C. \(x = 2y^2 - 40\)
• D. \(y = 2x^2 + 3x - 19\)
• E. \(x = 2y^2 + 3x - 19\)

Hint:

To identify a parabola equation from a graph, always check: 1. Direction (upward/downward). 2. Vertex location using \(-\tfrac{b}{2a}\). 3. Constant term from the \(y\)-intercept. This quickly eliminates wrong options.

Answer 1

The Correct Option is D

Step 1: Identify the type of curve.
From the graph, the curve is a parabola opening upwards. This means the general form is: \[ y = ax^2 + bx + c \] Step 2: Locate the vertex.
From the graph, the vertex occurs around \( (x, y) = ( -0.75, -20)\). So, the parabola has a minimum point close to \((-0.75, -20)\). Step 3: Use vertex formula.
The \(x\)-coordinate of the vertex is given by: \[ x_v = -\frac{b}{2a} \] Here, option (D) has \(a=2, b=3\).
So, \[ x_v = -\frac{3}{2(2)} = -\frac{3}{4} = -0.75 \] This matches the graph. Step 4: Find the \(y\)-coordinate of vertex.
Substitute \(x=-0.75\) into \(y = 2x^2 + 3x - 19\): \[ y = 2(-0.75)^2 + 3(-0.75) - 19 \] \[ = 2(0.5625) - 2.25 - 19 = 1.125 - 2.25 - 19 = -20.125 \approx -20 \] This matches the graph. Step 5: Verify other points.
At \(x=0\), \[ y = 2(0)^2 + 3(0) - 19 = -19 \] At \(x=4\), \[ y = 2(16) + 12 - 19 = 25 \] Both points match the graph. \[ \boxed{y = 2x^2 + 3x - 19} \]