Questions: Graph f(t) = t^2 + 3t - 4.

Graph f(t) = t^2 + 3t - 4.
Transcript text: Graph $f(t)=t^{2}+3 t-4$.
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Solution

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Solution Steps

Step 1: Identify the function

The given function is \( f(t) = t^2 + 3t - 4 \).

Step 2: Find the vertex of the parabola

The vertex form of a quadratic function \( at^2 + bt + c \) is given by \( t = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = 3 \), and \( c = -4 \).

\[ t = -\frac{3}{2 \cdot 1} = -\frac{3}{2} \]

Step 3: Calculate the value of the function at the vertex

Substitute \( t = -\frac{3}{2} \) into the function to find \( f(t) \).

\[ f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) - 4 \] \[ = \frac{9}{4} - \frac{9}{2} - 4 \] \[ = \frac{9}{4} - \frac{18}{4} - \frac{16}{4} \] \[ = \frac{9 - 18 - 16}{4} \] \[ = \frac{-25}{4} \] \[ = -6.25 \]

Step 4: Find the y-intercept

The y-intercept occurs when \( t = 0 \).

\[ f(0) = 0^2 + 3 \cdot 0 - 4 = -4 \]

Step 5: Find the x-intercepts

To find the x-intercepts, set \( f(t) = 0 \).

\[ t^2 + 3t - 4 = 0 \]

Solve the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \] \[ t = \frac{-3 \pm \sqrt{9 + 16}}{2} \] \[ t = \frac{-3 \pm \sqrt{25}}{2} \] \[ t = \frac{-3 \pm 5}{2} \]

So, the solutions are:

\[ t = \frac{-3 + 5}{2} = 1 \] \[ t = \frac{-3 - 5}{2} = -4 \]

Step 6: Plot the points and draw the parabola
  • Vertex: \((-1.5, -6.25)\)
  • Y-intercept: \((0, -4)\)
  • X-intercepts: \((1, 0)\) and \((-4, 0)\)

Final Answer

The graph of the function \( f(t) = t^2 + 3t - 4 \) is a parabola opening upwards with the vertex at \((-1.5, -6.25)\), y-intercept at \((0, -4)\), and x-intercepts at \((1, 0)\) and \((-4, 0)\).

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