Answer :
Sure, let's solve this problem step-by-step:
1. Sketch the graph of [tex]\( y = x^2 - 3x - 10 \)[/tex]:
- The equation [tex]\( y = x^2 - 3x - 10 \)[/tex] is a quadratic function, and its graph is a parabola opening upwards because the coefficient of [tex]\( x^2 \)[/tex] is positive.
- To sketch the graph, we first find the roots of the quadratic equation, i.e., the values of [tex]\( x \)[/tex] for which [tex]\( y = 0 \)[/tex].
2. Find the roots of the equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex]:
- Solving [tex]\( x^2 - 3x - 10 = 0 \)[/tex], we find the roots to be:
[tex]\[ x_1 = -2 \quad \text{and} \quad x_2 = 5 \][/tex]
These roots indicate the points where the parabola intersects the [tex]\( x \)[/tex]-axis.
3. Sketch the graph:
- The parabola passes through the points [tex]\( (-2, 0) \)[/tex] and [tex]\( (5, 0) \)[/tex].
- The vertex of the parabola can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex] and [tex]\( b = -3 \)[/tex]:
[tex]\[ x = \frac{3}{2} = 1.5 \][/tex]
Plug this value back into the equation to get the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ y = (1.5)^2 - 3 \cdot 1.5 - 10 = -\frac{49}{4} \][/tex]
- So, the vertex of the parabola is [tex]\( \left(1.5, -\frac{49}{4}\right) \)[/tex].
- Using these points, you can sketch the parabola which opens upwards, with intercepts at [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex].
4. Identify the interval where [tex]\( x^2 - 3x - 10 < 0 \)[/tex]:
- The expression [tex]\( x^2 - 3x - 10 < 0 \)[/tex] holds true between the roots [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex], because this is the interval where the parabola is below the [tex]\( x \)[/tex]-axis.
5. Determine the integer values within this interval:
- The integer values of [tex]\( x \)[/tex] that lie strictly between [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex] are:
[tex]\[ -1, 0, 1, 2, 3, 4 \][/tex]
Therefore, the integer values that satisfy [tex]\( x^2 - 3x - 10 < 0 \)[/tex] are:
[tex]\[ -1, 0, 1, 2, 3, 4 \][/tex]
1. Sketch the graph of [tex]\( y = x^2 - 3x - 10 \)[/tex]:
- The equation [tex]\( y = x^2 - 3x - 10 \)[/tex] is a quadratic function, and its graph is a parabola opening upwards because the coefficient of [tex]\( x^2 \)[/tex] is positive.
- To sketch the graph, we first find the roots of the quadratic equation, i.e., the values of [tex]\( x \)[/tex] for which [tex]\( y = 0 \)[/tex].
2. Find the roots of the equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex]:
- Solving [tex]\( x^2 - 3x - 10 = 0 \)[/tex], we find the roots to be:
[tex]\[ x_1 = -2 \quad \text{and} \quad x_2 = 5 \][/tex]
These roots indicate the points where the parabola intersects the [tex]\( x \)[/tex]-axis.
3. Sketch the graph:
- The parabola passes through the points [tex]\( (-2, 0) \)[/tex] and [tex]\( (5, 0) \)[/tex].
- The vertex of the parabola can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex] and [tex]\( b = -3 \)[/tex]:
[tex]\[ x = \frac{3}{2} = 1.5 \][/tex]
Plug this value back into the equation to get the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ y = (1.5)^2 - 3 \cdot 1.5 - 10 = -\frac{49}{4} \][/tex]
- So, the vertex of the parabola is [tex]\( \left(1.5, -\frac{49}{4}\right) \)[/tex].
- Using these points, you can sketch the parabola which opens upwards, with intercepts at [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex].
4. Identify the interval where [tex]\( x^2 - 3x - 10 < 0 \)[/tex]:
- The expression [tex]\( x^2 - 3x - 10 < 0 \)[/tex] holds true between the roots [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex], because this is the interval where the parabola is below the [tex]\( x \)[/tex]-axis.
5. Determine the integer values within this interval:
- The integer values of [tex]\( x \)[/tex] that lie strictly between [tex]\( -2 \)[/tex] and [tex]\( 5 \)[/tex] are:
[tex]\[ -1, 0, 1, 2, 3, 4 \][/tex]
Therefore, the integer values that satisfy [tex]\( x^2 - 3x - 10 < 0 \)[/tex] are:
[tex]\[ -1, 0, 1, 2, 3, 4 \][/tex]
