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Math  /  Algebra

QuestionGraph f(t)=t2+10t+23f(t)=t^{2}+10 t+23. You ony need to graph the vertext and one other point.

Studdy Solution

STEP 1

1. The function f(t)=t2+10t+23 f(t) = t^2 + 10t + 23 is a quadratic function.
2. The graph of a quadratic function is a parabola.
3. The vertex of the parabola can be found using the vertex formula.
4. We need to find the vertex and one additional point to sketch the graph.

STEP 2

1. Identify the coefficients of the quadratic function.
2. Calculate the vertex of the parabola.
3. Find one additional point on the parabola.
4. Sketch the graph using the vertex and the additional point.

STEP 3

Identify the coefficients of the quadratic function:
The given function is f(t)=t2+10t+23 f(t) = t^2 + 10t + 23 . The coefficients are: - a=1 a = 1 (coefficient of t2 t^2 ) - b=10 b = 10 (coefficient of t t ) - c=23 c = 23 (constant term)

STEP 4

Calculate the vertex of the parabola using the vertex formula:
The vertex (t,f(t)) (t, f(t)) of a parabola given by f(t)=at2+bt+c f(t) = at^2 + bt + c can be found using the formula:
t=−b2a t = -\frac{b}{2a}
Substitute the values of a a and b b :
t=−102×1=−102=−5 t = -\frac{10}{2 \times 1} = -\frac{10}{2} = -5
Now, calculate f(−5) f(-5) :
f(−5)=(−5)2+10(−5)+23 f(-5) = (-5)^2 + 10(-5) + 23 f(−5)=25−50+23 f(-5) = 25 - 50 + 23 f(−5)=−2 f(-5) = -2
Thus, the vertex is (−5,−2) (-5, -2) .

STEP 5

Find one additional point on the parabola:
Choose a value for t t that is easy to compute, such as t=0 t = 0 :
f(0)=02+10(0)+23 f(0) = 0^2 + 10(0) + 23 f(0)=23 f(0) = 23
Thus, the additional point is (0,23) (0, 23) .

STEP 6

Sketch the graph using the vertex and the additional point:
1. Plot the vertex (−5,−2) (-5, -2) on the graph.
2. Plot the additional point (0,23) (0, 23) .
3. Draw a smooth curve through these points, ensuring the parabola opens upwards (since a=1>0 a = 1 > 0 ).
The graph of f(t)=t2+10t+23 f(t) = t^2 + 10t + 23 has its vertex at (−5,−2) (-5, -2) and passes through the point (0,23) (0, 23) .

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