Primeira página
/
Matemática
/
25. f(x)=2x^2-x-1 (a) Is the point (-1,2) on the graph of f? (b) If x=-2 what is f(x) ? What point is on the graph of f? (c) If f(x)=-1 , what is x? What point (s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept , if there is one of the graph of f.

Pergunta

25. f(x)=2x^2-x-1
(a) Is the point (-1,2) on the graph of f?
(b) If x=-2 what is f(x) ? What point is on the graph of f?
(c) If f(x)=-1 , what is x? What point (s) are on the graph
of f?
(d) What is the domain of f?
(e) List the x-intercepts, if any, of the graph of f.
(f) List the y-intercept , if there is one of the graph of f.

25. f(x)=2x^2-x-1 (a) Is the point (-1,2) on the graph of f? (b) If x=-2 what is f(x) ? What point is on the graph of f? (c) If f(x)=-1 , what is x? What point (s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept , if there is one of the graph of f.

Solução

expert verifiedVerification of experts
4.2279 Voting
avatar
ArthurElite · Tutor por 8 anos

Responder

Let's address each part of the question step by step.<br /><br />### (a) Is the point \((-1, 2)\) on the graph of \( f \)?<br /><br />To determine if the point \((-1, 2)\) is on the graph of \( f(x) = 2x^2 - x - 1 \), we need to check if \( f(-1) = 2 \).<br /><br />Calculate \( f(-1) \):<br />\[ f(-1) = 2(-1)^2 - (-1) - 1 \]<br />\[ f(-1) = 2(1) + 1 - 1 \]<br />\[ f(-1) = 2 + 1 - 1 \]<br />\[ f(-1) = 2 \]<br /><br />Since \( f(-1) = 2 \), the point \((-1, 2)\) is indeed on the graph of \( f \).<br /><br />### (b) If \( x = -2 \), what is \( f(x) \)? What point is on the graph of \( f \)?<br /><br />Calculate \( f(-2) \):<br />\[ f(-2) = 2(-2)^2 - (-2) - 1 \]<br />\[ f(-2) = 2(4) + 2 - 1 \]<br />\[ f(-2) = 8 + 2 - 1 \]<br />\[ f(-2) = 9 \]<br /><br />So, when \( x = -2 \), \( f(x) = 9 \). The point \((-2, 9)\) is on the graph of \( f \).<br /><br />### (c) If \( f(x) = -1 \), what is \( x \)? What point(s) are on the graph of \( f \)?<br /><br />We need to solve the equation \( 2x^2 - x - 1 = -1 \):<br />\[ 2x^2 - x - 1 + 1 = 0 \]<br />\[ 2x^2 - x = 0 \]<br />\[ x(2x - 1) = 0 \]<br /><br />This gives us two solutions:<br />\[ x = 0 \quad \text{or} \quad 2x - 1 = 0 \]<br />\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]<br /><br />So, the values of \( x \) are \( 0 \) and \( \frac{1}{2} \). The points on the graph are \((0, -1)\) and \(\left(\frac{1}{2}, -1\right)\).<br /><br />### (d) What is the domain of \( f \)?<br /><br />The function \( f(x) = 2x^2 - x - 1 \) is a polynomial, and polynomials are defined for all real numbers. Therefore, the domain of \( f \) is:<br />\[ \text{Domain of } f: (-\infty, \infty) \]<br /><br />### (e) List the x-intercepts, if any, of the graph of \( f \).<br /><br />To find the x-intercepts, we set \( f(x) = 0 \):<br />\[ 2x^2 - x - 1 = 0 \]<br /><br />We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -1 \):<br /><br />\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} \]<br />\[ x = \frac{1 \pm \sqrt{1 + 8}}{4} \]<br />\[ x = \frac{1 \pm \sqrt{9}}{4} \]<br />\[ x = \frac{1 \pm 3}{4} \]<br /><br />This gives us two solutions:<br />\[ x = \frac{1 + 3}{4} = 1 \]<br />\[ x = \frac{1 - 3}{4} = -\frac{1}{2} \]<br /><br />So, the x-intercepts are \( x = 1 \) and \( x = -\frac{1}{2} \). The points are \((1, 0)\) and \(\left(-\frac{1}{2}, 0\right)\).<br /><br />### (f) List the y-intercept, if there is one, of the graph of \( f \).<br /><br />To find the y-intercept, we set \( x = 0 \):<br />\[ f(0) = 2(0)^2 - 0 - 1 \]<br />\[ f(0) = -1 \]<br /><br />So, the y-intercept is \( y = -1 \). The point is \((0, -1)\).<br /><br />In summary:<br />(a) Yes, \((-1, 2)\) is on the graph.<br />(b) When \( x = -2 \), \( f(x) = 9 \). The point is \((-2, 9)\).<br />(c) When \( f(x) = -1 \), \( x = 0 \) or \( x = \frac{1}{2} \). The points are \((0, -1)\) and \(\left(\frac{1}{2}, -1\right)\).<br />(d) The domain of \( f \) is \((-\infty, \infty)\).<br />(e) The x-intercepts are \((1, 0)\) and \(\left(-\frac{1}{2}, 0\right)\).<br />(f) The y-intercept is \((0, -1)\).
Clique para avaliar: